Initial Value Problems Question Bank Set 2

MATH 334 - DIFFERENTIAL EQUATIONS - Initial value problems Question Bank - Set 2 Liberty University Question 1 Question Solve the initial value problem: dy = y 2 − 4y, dt y(0) = 3 Solution Step 1: Separate variables: dy = y 2 − 4y dt y2 dy = dt − 4y dy = dt y(y − 4) Step 2: Partial fraction decomposition: 1 A B = + y(y − 4) y y−4 1 = A(y − 4) + By Putting y = 0 gives 1 = −4A, so A = − 14 . Putting y = 4 gives 1 = 4B, so B = 41 . Therefore, the partial fraction decomposition is: 1 1 1 =− + y(y − 4) 4y 4(y − 4) Step 3: Integrate both sides: Z Z 1 dy = dt y(y − 4)  Z  Z 1 1 dy = dt − + 4y 4(y − 4) 1 1 − ln |y| + ln |y − 4| = t + C 4 4 y−4 = 4t + C ln y y−4 = ke4t y where k = eC . Step 4: Apply initial condition: Substitute y = 3 and t = 0 into the solution to find k: 3−4 1 =k·1 ⇒ k =− 3 3 Step 5: Final solution: Thus, the solution to the initial value problem is: y−4 1 = − e4t y 3 y= 4 1 + 3e4t Question 2 Question Consider the initial value problem: dy = 2y − x, y(0) = 1 dx Find the particular solution to the initial value problem. Solution Step 1: First, we need to solve the differential equation. dy = 2y − x Given: dx Separating variables, we get: dy = dx 2y − x Integrating both sides gives: 2 Z 1 dy = 2y − x Z dx Step 2: Continuing from the previous step, we solve the integrals: Let u = 2y − x, then du = 2dy Therefore, the integral on the left side becomes: Z 1 1 1 du = ln |u| + C1 2 u 2 And the integral on the right side is simply: Z dx = x + C2 Step 3: Next, we substitute back for u and simplify: 1 ln |2y − x| = x + C 2 Where C = C2 − C1 is a constant of integration. Step 4: Now, we apply the initial condition y(0) = 1 to find the particular solution. When x = 0 and y = 1, we have: 1 ln |2(1) − 0| = 0 + C 2 1 ln 2 = C 2 So, the particular solution is: 1 1 ln |2y − x| = x + ln 2 2 2 Question 3 Question Consider the initial value problem: dy = 2y − x2 , dx y(0) = 1. Find the particular solution of the differential equation that satisfies the initial condition. 3 Solution Step 1: First, let’s solve the differential equation by separating variables. We have: dy = 2y − x2 . dx Step 2: Rearranging the terms, we get: dy = dx. 2y − x2 Step 3: Now, integrate both sides with respect to y and x separately. Z Z 1 dy = dx. 2y − x2 Step 4: The integral on the left side can be simplified by using substitution. Let u = 2y − x2 , then du = 2 dy. Z Z 1 1 du = dx. 2 u Step 5: Integrating, we have: 1 ln |u| = x + C, 2 where C is the constant of integration. Step 6: Substitute back u = 2y − x2 into the equation: 1 ln |2y − x2 | = x + C. 2 Step 7: Since y(0) = 1, substitute x = 0 and y = 1 into the equation to solve for C: 1 ln |2(1) − 02 | = 0 + C. 2 1 ln |2| = C. 2 C = ln |2|1/2 . Step 8: Finally, substitute the value of C back into the equation to find the particular solution: 1 ln |2y − x2 | = x + ln |2|1/2 . 2 Therefore, the particular solution of the differential equation that satisfies the initial condition y(0) = 1 is 21 ln |2y − x2 | = x + ln |2|1/2 . 4 Question 4 Question x Solve the initial value problem: y ′ = 12 y + 3e 2 , y(0) = 1. Solution Step 1: First, we find the integrating factorR for the given differential equation. x 1 The integrating factor is given by µ(x) = e 2 dx = e 2 . Step 2: Multiply both sides of the differential equation by the integrating factor: x 1 x e 2 y′ − e 2 y = 3 2 Step 3: Notice that the left side is the result of applying the product rule to x e 2 y. So we can rewrite the equation as: d x (e 2 y) = 3 dx Step 4: Integrate both sides with respect to x: Z Z d x (e 2 y)dx = 3dx dx x e 2 y = 3x + C Step 5: Apply the initial condition y(0) = 1 to find the value of C: e0 · 1 = 3 · 0 + C C=1 Step 6: Substitute the value of C back into the equation found in Step 4: x e 2 y = 3x + 1 Step 7: Solve for y: x x y = 3xe− 2 + e− 2 x x Therefore, the solution to the initial value problem is y = 3xe− 2 + e− 2 . Question 5 Question Solve the initial value problem: dy dx + 2xy = x, y(0) = 1. 5 Solution To solve this differential equation, we will use an integrating factor method. Step 1: Identify the integrating factor. The integrating factor µ(x) is given by: R µ(x) = e 2x dx Step 2: Find the integrating factor. Z 2x dx = x2 Hence, the integrating factor is: µ(x) = ex 2 Step 3: Multiply the differential equation by the integrating factor. The differential equation becomes: 2 ex 2 2 dy + 2xex y = xex dx Step 4: Rewrite the left-hand side as the derivative of a product. Using the product rule, we have: 2 d  x2  e y = xex dx Step 5: Integrate both sides. Integrating both sides gives: Z 2 2 ex y = xex dx + C 1 x2 e +C 2 Step 6: Solve for y. Solving for y, we get: 2 ex y = y= 2 1 + Ce−x 2 Step 7: Apply the initial condition. Using the initial condition y(0) = 1, we find: 1 1= +C 2 1 C= 2 Step 8: Final solution. Therefore, the solution to the initial value problem is: 2 1 1 y = + e−x 2 2 6 Question 6 Question Consider the initial value problem y ′ − 2y = e2x , where y(0) = 1. Find the particular solution to the differential equation that satisfies the initial condition. Solution Step 1: First, we find the integrating factor I(x). Let y ′ − 2y = e2x . The integrating factor is given by I(x) = e R −2dx = e−2x . Step 2: Multiply both sides of the differential equation by the integrating factor and simplify. Multiply e−2x to both sides of the equation to get e−2x (y ′ − 2y) = e−2x e2x . This simplifies to d −2x (e y) = 1. dx Step 3: Integrate both sides of the equation. Integrating both sides with respect to x, we get Z Z d −2x (e y)dx = 1dx. dx So, we have e−2x y = x + C, where C is the constant of integration. Step 4: Apply the initial condition y(0) = 1 to solve for the constant C. Substitute x = 0 and y = 1 into the equation above: e0 · 1 = 0 + C. Therefore, C = 1. Step 5: Finally, find the particular solution. Substitute the value of C back into the equation above: e−2x y = x + 1. Hence, the particular solution to the initial value problem is given by y = e2x (x + 1). Question 7 Question Solve the initial value problem: dy = −2x(y + 1), dx 7 y(0) = 2 Solution Step 1: We start by separating the variables in the given differential equation: dy = −2xdx y+1 Step 2: Next, we integrate both sides of the equation: Z Z 1 dy = −2xdx y+1 Step 3: Integrating the left side with respect to y and the right side with respect to x, we get: ln |y + 1| = −x2 + C where C is the constant of integration. Step 4: Exponentiating both sides of the equation yields: |y + 1| = e−x 2 +C = eC e−x 2 Step 5: Since y(0) = 2, we substitute x = 0 and y = 2 into the equation to find C: |2 + 1| = eC ⇒ 3 = eC ⇒ C = ln 3 Step 6: Therefore, the solution to the initial value problem is: 2 y = e−x − 1 Question 8 Question Consider the initial value problem: dy = 2x2 − 4x + 5, dx y(0) = 3. Find the solution to the initial value problem. Solution Step 1: Find the general solution dy = 2x2 − 4x + 5. dx Z Z dy = (2x2 − 4x + 5)dx. y= 2 3 x − 2x2 + 5x + C. 3 8 where C is the constant of integration. Step 2: Use the initial condition to find the particular solution y(0) = 3, 2 3 (0) − 2(0)2 + 5(0) + C = 3, 3 C = 3. Therefore, the solution to the initial value problem is: y= 2 3 x − 2x2 + 5x + 3. 3 Question 9 Question Solve the initial value problem: y ′ = sin(2x) − 2y, y(0) = 1 Solution Step 1: First, we find the general solution to the homogeneous differential equation y ′ = −2y by separating variables: dy dy = −2y ⇒ = −2dx ⇒ ln |y| = −2x + C1 dx y where C1 is the constant of integration. Exponentiating both sides gives |y| = eC1 e−2x . Since eC1 is just another constant, we can simplify this to |y| = Ce−2x , where C is a constant. Step 2: Now, we will find a particular solution to the non-homogeneous part of the equation. Let yp be a particular solution to the equation y ′ = sin(2x). We can guess yp = A sin(2x) + B cos(2x) and find A and B by substitution: yp′ = 2A cos(2x) − 2B sin(2x) = sin(2x) Comparing coefficients, we get 2A = 0 and −2B = 1, which gives A = 0 and B = − 21 . Therefore, yp = − 12 cos(2x). Step 3: The general solution to the non-homogeneous differential equation is the sum of the particular solution and the general solution to the homogeneous equation, so y = Ce−2x − 12 cos(2x). Step 4: Using the initial condition y(0) = 1, we find: 1 = Ce0 − 1 3 1 cos(0) ⇒ 1 = C − ⇒ C = 2 2 2 Therefore, the solution to the initial value problem is y = 32 e−2x − 12 cos(2x). 9 Question 10 Question Consider the initial value problem: dy 3x2 + y = , dx 2x y(1) = 4 Find the particular solution to the initial value problem. Solution Step 1: Rewrite the given differential equation in a more convenient form. dy 3x2 + y dy = =⇒ 2x = 3x2 + y dx 2x dx Step 2: Recognize that the differential equation is linear and in standard form y ′ + P (x)y = Q(x), where P (x) = x2 and Q(x) = 3x2 . R P (x)dx Step 3: Find the integrating factor µ(x) = e . µ(x) = e R 2 x dx 2 = e2 ln |x| = eln x = x2 Step 4: Multiply the integrating factor by both sides of the differential equation. dy x2 · − 2xy = 3x3 dx Step 5: Rewrite the equation as the derivative of a product. d 2 (x y) = 3x3 dx Step 6: Integrate both sides of the equation with respect to x. Z Z d(x2 y) = 3x3 dx Step 7: Solve the integrals. x2 y = 3 4 x +C 4 Step 8: Apply the initial condition y(1) = 4 to find the value of the constant C. 3 4 3 16 3 13 (1) + C =⇒ 4 = + C =⇒ C = − = 4 4 4 4 4 Step 9: Write the particular solution. (1)2 · 4 = x2 y = 3 4 13 3 13 x + =⇒ y = x2 + 2 4 4 4 4x Therefore, the particular solution to the initial value problem is y = 34 x2 + 13 4x2 . 10 Question 11 Question Consider the initial value problem: dy 2x − y = , dx 2y − x y(0) = 1 Solve the initial value problem. Solution To solve the initial value problem, we will first rewrite the given differential equation in a more familiar form by letting u = xy . Step 1: Rewrite the differential equation using u = y x Let u = xy . dy = x du Then, we have y = ux and dx dx + u. y Substitute u = x into the given differential equation to obtain: x du 2x − ux +u= dx 2ux − x Step 2: Solve for u Solving the above equation for u gives: x x x du 2x − ux +u= dx 2ux − x 2x(1 − u) du +u= dx x(2u − 1) x 2(1 − u) du +u= dx 2u − 1 x du 2(1 − u) = −u dx 2u − 1 2 − 2u − 2u(2u − 1) du = dx 2u − 1 x du 2 − 2u − 4u2 + 2u = dx 2u − 1 du 2 − 4u2 = dx 2u − 1 Separate variables, x and u, to solve for u by integrating: x 1 (2u − 1)du = (2 − 4u2 ) dx x 11 Question 12 Question Solve the initial value problem: y′ = 1 , x2 + y y(1) = −1. Solution Step 1: Rewrite the differential equation in standard form. 1 dy = 2 dx x +y Step 2: Separate variables to solve the differential equation. (x2 + y) dy = dx Step 3: Integrate both sides of the equation. Z Z 2 (x + y) dy = dx 1 x2 y + y 2 = x + C 2 Step 4: Use the initial condition y(1) = −1 to find the value of C. 1 12 (−1) + (−1)2 = 1 + C 2 −1 − 1 =1+C 2 3 =1+C 2 5 C=− 2 Step 5: Substitute the value of C back into the equation found in Step 3. − 1 5 x2 y + y 2 = x − 2 2 Step 6: Now, we need to solve for y. This is a nonlinear equation, and solving for y explicitly may be difficult. Therefore, we have found the general solution implicitly. 12 Question 13 Question Find the particular solution to the initial value problem: y ′ = 2x + 3e2y , y(0) = 1. Solution Step 1: First, separate the variables in the given differential equation: dy = 2x + 3e2y dx Step 2: Next, integrate both sides with respect to x: Z Z Z dy = 2xdx + 3e2y dx Step 3: Integrate each term separately: y = x2 + C1 + 3 Z e2y dx Step 4: To integrate e2y with respect to y, let u = 2y so du = 2dy: Z 3 u e du y = x2 + C1 + 2 Step 5: Evaluate the integral: 3 y = x2 + C1 + e2y + C2 2 Step 6: Simplify the equation: 3 y − e2y = x2 + C 2 Step 7: Use the initial condition y(0) = 1 to find the value of the constant C: 3 1 − e2 = 0 + C 2 Step 8: Solve for C: 3 2 e −1 2 Step 9: Substitute the value of C back into the equation to find the particular solution: 3 3 y − e2y = x2 + e2 − 1 2 2 Therefore, the particular solution to the initial value problem is y = x2 + 3 2 3 2y 2e − 1 + 2e . C= 13 Question 14 Question Solve the initial value problem: dy = t2 − 2y, dt y(0) = 1 Solution Step 1: First, let’s find the general solution to the differential equation dy dt = dy 2 2 t − 2y. We can rewrite this equation as dt + 2y = t . Step 2: This is a first-order linear differential equation, so we can use the R integrating factor method. The integrating factor is e 2 dt = e2t . Step 3: Multiply the integrating factor by the equation to get: e2t dy + 2e2t y = e2t t2 dt Step 4: Rewrite the left side as a derivative of a product: d 2t (e y) = e2t t2 dt Step 5: Integrate both sides with respect to t: Z Z d 2t (e y) dt = e2t t2 dt dt Step 6: Integrating gives: e2t y = Z t2 e2t dt Step 7: Now, we need to solve the integral on the right side. This requires integrating by parts one or more times. Step 8: After integrating by parts, we find: Z Z 1 2 2t 1 2 2t t e dt = t e − e2t dt 2 2 Step 9: Continuing to solve the integral, we have: Z 1 e2t dt = e2t + C 2 Step 10: Substituting this back in, we have: e2t y = 1 2 2t 1 2t C t e − e − 2 4 2 14 Step 11: Solve for y by dividing by e2t : y= C 1 2 1 t − − 2t 2 4 2e Step 12: Use the initial condition y(0) = 1 to find the value of C. Substitute t = 0 and y = 1 into the equation. Step 13: Solving for C gives C = −1/4. Step 14: Substitute C back into the equation for y to obtain the final solution. Therefore, the solution to the initial value problem is: y= 1 2 1 1 t − − 2t 2 4 2e Question 15 Question Find the particular solution to the initial value problem: dy = 3x2 + 4x, dx y(0) = 5. Solution Step 1: First, we solve the differential equation to find the general solution. dy = 3x2 + 4x dx Integrating both sides with respect to x, we get: Z Z dy = (3x2 + 4x)dx y = x3 + 2x2 + C where C is a constant. Step 2: Next, we apply the initial condition y(0) = 5 to find the particular solution. 5 = 03 + 2(0)2 + C C=5 Step 3: Therefore, the particular solution to the initial value problem is: y = x3 + 2x2 + 5 15 Question 16 Question Solve the initial value problem: dy = 2xy, dx y(0) = −1. Solution Step 1: Write the initial value problem as a differential equation. dy = 2xy, y(0) = −1. dx Step 2: Separate variables to solve the differential equation. dy = 2xdx. y Step 3: Integrate both sides. Z 1 dy = y Z 2xdx. Step 4: Solve the integrals. ln |y| = x2 + C, where C is the constant of integration. Step 5: Solve for y. 2 2 y = ex +C = Cex . Step 6: Use the initial condition y(0) = −1 to find the value of C. −1 = C. Step 7: Substitute C = −1 back into the equation for y. 2 y = −ex . 2 Therefore, the solution to the initial value problem is y = −ex . Question 17 Question Solve the initial value problem: dy 1 = y dx 2 with y(0) = 3 16 Solution Step 1: First, we rearrange the differential equation to separate variables: dy 1 = dx y 2 Step 2: Next, we integrate both sides: Z Z 1 1 dy = dx y 2 Step 3: Integrating, we get: ln |y| = 1 x+C 2 Step 4: Using the initial condition y(0) = 3, we find the value of the constant C: ln |3| = 0 + C =⇒ C = ln(3) Step 5: Substituting this value back into the equation, we have: ln |y| = 1 x + ln(3) 2 Step 6: Exponentiating both sides to eliminate the natural logarithm: 1 1 y = e 2 x+ln(3) = 3e 2 x 1 Therefore, the solution to the initial value problem is y = 3e 2 x . Question 18 Question Consider the initial value problem: dy = e2x sin(3x), dx y(0) = 1 Find the particular solution to the initial value problem. Solution Step 1: First, we need to solve the differential equation grating both sides with respect to x gives: Z Z dy dx = e2x sin(3x) dx dx 17 dy dx = e2x sin(3x). Inte- 1 2x e (3 sin(3x) − cos(3x)) + C 13 Step 2: Next, we use the initial condition y(0) = 1 to find the value of C. Substitute x = 0 and y = 1 into the equation: y=− 1=− 1 (3 sin(0) − cos(0)) + C 13 1 (−1) + C 13 1 12 1= +C ⇒C = 13 13 12 Step 3: Finally, substitute C = 13 back into the general solution to find the particular solution: 1=− y=− 12 1 2x e (3 sin(3x) − cos(3x)) + 13 13 1 2x Therefore, the particular solution to the initial value problem is y = − 13 e (3 sin(3x)− 12 cos(3x)) + 13 . Question 19 Question Solve the initial value problem: dy dx = e−2y , y(0) = 1. Solution To solve this initial value problem, we will separate variables and then integrate both sides. Step 1: Separate variables by writing the differential equation as follows: dy = e−2y dx e2y dy = dx Step 2: Integrate both sides: Z e2y dy = Z dx 1 2y e =x+C 2 where C is the constant of integration. Step 3: Solve for y using the initial condition y(0) = 1: 1 2·1 e =0+C 2 18 1 2 e =C 2 Step 4: Substitute back into the equation to find the solution: 1 2y 1 e = x + e2 2 2 1 e2y = 2(x + e2 ) 2
2y = ln 2x + e2 y=
1 ln 2x + e2 2 Therefore, the solution to the initial value problem is y = 1 2
ln 2x + e2 . Question 20 Question Consider the initial value problem: y′ = 2y + sin(t), t y(1) = 1. Find the particular solution to the initial value problem. Solution Step 1: First, we rewrite the given initial value problem in standard form by separating variables: dy 2y = + sin(t). dt t Step 2: Next, we isolate y terms and t terms: dt dy = + sin(t)dt. 2y t Step 3: We integrate both sides: Z Z Z 1 1 dy = dt + sin(t) dt. 2y t Step 4: Solving the integrals gives us: 1 ln |y| = ln |t| − cos(t) + C, 2 where C is the constant of integration. 19 Step 5: Simplifying and solving for y, we have: ln |y| = 2 ln |t| − 2 cos(t) + C ′ . Here, C ′ = 2C. Step 6: Exponentiating both sides, we obtain: |y| = e2 ln |t| · e−2 cos(t)+C ′ ′ |y| = t2 e−2 cos(t)+C . Step 7: Applying the initial condition y(1) = 1, we get: ′ |1| = 12 e−2 cos(1)+C , which gives us C ′ = 0. Step 8: Therefore, the particular solution to the initial value problem is: y = t2 e−2 cos(t) . Question 21 Question Consider the initial value problem dy = 2y − x, dx y(0) = 1. Find the particular solution y(x) satisfying the initial condition. Solution Step 1: First, let’s solve the differential equation using the method of integrating R factor. The integrating factor is given by µ(x) = e 2 dx = e2x . Multiplying both sides of the differential equation by the integrating factor, we have e2x dy = e2x (2y − x). dx Step 2: Notice that the left side can be written as the derivative of the product e2x y. Therefore, we have d 2x (e y) = e2x (2y − x). dx Step 3: Integrating both sides with respect to x, we get Z 2x e y = e2x (2y − x) dx. 20 Step 4: Solving the integral on the right side, we have Z Z 2x 2x e y = 2 e y dx − xe2x dx. R Step 5: Integrating xe2x dx by parts, let u = x and dv = e2x dx, then du = dx and v = 21 e2x . Hence, we have 1 1 e2x y = 2e2x y − xe2x − e2x + C, 2 4 where C is the constant of integration. Step 6: Simplifying the equation, we get 1 1 e2x y − 2e2x y = − xe2x − e2x + C. 2 4 Step 7: Continuing to simplify, we find 1 1 −y = − x − + Ce−2x . 2 4 Step 8: Finally, applying the initial condition y(0) = 1, we get −1 = 0 − 1 +C 4 3 C=− . 4 ⇒ Step 9: Therefore, the particular solution to the initial value problem is 1 1 3 y(x) = − x − − e−2x . 2 4 4 Question 22 Question Solve the initial value problem: dy = 2x2 − 3x, dx y(1) = 4 Solution Step 1: Find the general solution by solving the differential equation. dy = 2x2 − 3x dx Integrating both sides with respect to x, we get: Z y = (2x2 − 3x)dx 21 2 3 3 2 x − x +C 3 2 where C is the constant of integration. Step 2: Use the initial condition y(1) = 4 to find the value of C. y= 4= 2 3 3 2 (1) − (1) + C 3 2 2 3 2 4= 3 2 4= 3 4 4= 6 3 +C 2 9 − +C 6 3 − +C 6 3 − +C 6 1 4= +C 6 1 C =4− 6 23 C= 6 Step 3: Substitute the value of C back into the general solution. 4= y= − 2 3 3 2 23 x − x + 3 2 6 Therefore, the solution to the initial value problem is: y= 2 3 3 2 23 x − x + 3 2 6 Question 23 Question Solve the initial value problem: dy = 3y − 2x, dx 22 y(0) = 4 Solution Step 1: First, we write the given differential equation in the standard form dy dx − 3y = −2x. R Step 2: Next, we find the integrating factor by computing e −3 dx . Integrating factor = e−3x Step 3: We multiply the integrating factor with both sides of the differential equation: dy − 3e−3x y = −2xe−3x e−3x dx Step 4: The left-hand side of the equation can be rewritten using the product d (ye−3x ). rule as dx d (ye−3x ) = −2xe−3x dx Step 5: Now, we integrate both sides with respect to x: Z Z d (ye−3x ) dx = −2xe−3x dx dx Step 6: Integrating both sides gives us: ye−3x = 2 2 −3x x e +C 9 where C is the constant of integration. Step 7: To find the value of C, we use the initial condition y(0) = 4. ⇒ 4=0+C C=4 Step 8: Substituting the value of C back into the equation, we have: ye−3x = 2 2 −3x x e +4 9 Step 9: Finally, solving for y gives: y= 2 2 x + 4e3x 9 Therefore, the solution to the initial value problem is y = 29 x2 + 4. Question 24 Question Consider the initial value problem: dy = x2 + y, dx y(0) = 2 Find the particular solution to the initial value problem. 23 Solution Step 1: First, let’s rewrite the given initial value problem into a more familiar form: dy − y = x2 , y(0) = 2 dx dy + Step 2: This is a first-order linear differential equation of the form dx 2 P (x)y = Q(x), where P (x) = −1 and Q(x) = x . Step 3: To solveR the differential equation, we Rmultiply both sides by the integrating factor, e P (x)dx , which in this case is e −1dx = e−x . This gives us: e−x dy − e−x y = e−x x2 dx Step 4: Notice that the left-hand side is equal to rule. Therefore, we rewrite the equation as: d −x ) dx (ye by the product d (ye−x ) = e−x x2 dx Step 5: Now we integrate both sides with respect to x to solve for y: Z Z d −x (ye )dx = e−x x2 dx dx Step 6: Integrating both sides gives us: ye−x = −e−x x2 − 2e−x + C where C is the constant of integration. Step 7: To solve for y, multiply through by ex and simplify: y = −x2 − 2 + Cex Step 8: Using the initial condition, y(0) = 2, we can find the value of the constant C: 2 = −02 − 2 + Ce0 2 = −2 + C ⇒ C = 4 Step 9: Therefore, the particular solution to the initial value problem is: y = −x2 − 2 + 4ex Question 25 Question Solve the initial value problem: dy = 2x + 3y, dx 24 y(0) = 1 Solution Step 1: First, let’s rewrite the given initial value problem as: dy − 3y = 2x dx with y(0) = 1 Step 2: This is a first-order linear differential equation. We can solve it by finding the integrating factor. The integrating factor I(x) is given by: I(x) = e R −3dx = e−3x Step 3: Now, multiply both sides of the differential equation by the integrating factor: dy − 3e−3x y = 2xe−3x e−3x dx Step 4: The left side can be written in the form of the product rule: d −3x (e y) = 2xe−3x dx Step 5: Integrate both sides with respect to x: Z Z −3x d(e y) = 2xe−3x dx Step 6: This gives us: Z 2 2 e−3x y = 2xe−3x dx = − xe−3x − e−3x + C 3 9 Step 7: Solving for y, we get: 2 2 y = − x − + Ce3x 3 9 Step 8: Using the initial condition y(0) = 1, we can find the value of the constant C: 2 11 1=− +C ⇒ C = 9 9 Step 9: Therefore, the solution to the initial value problem is: 2 2 11 y = − x − + e3x 3 9 9 Question 26 Question Solve the initial value problem: y′ = 2x + y , x + 2y 25 y(0) = 1 Solution To solve the initial value problem, we will use separation of variables. Let’s begin by rewriting the differential equation: y′ = 2x + y x + 2y Step 1: Write the equation in the form g(y) dy = f (x) dx: (x + 2y) dy = (2x + y) dx Step 2: Integrate both sides: Z Z (x + 2y) dy = (2x + y) dx Z Z x dy + Z 2y dy = Z 2x dx + y dx xy + y 2 = x2 + y + C Step 3: Use the initial condition y(0) = 1 to solve for the constant C: 0 ∗ 1 + 12 = 02 + 1 + C C=1 Step 4: Substitute C = 1 back into the equation: xy + y 2 = x2 + y + 1 Step 5: Since we cannot solve explicitly for y, we have found the implicit solution to the differential equation: xy + y 2 = x2 + y + 1 Question 27 Question Consider the initial value problem: dy = y 2 + 2x, dx y(0) = 1 Find the solution to the initial value problem. 26 Solution Step 1: Rewrite the initial value problem as a separable differential equation: dy = y 2 + 2x dx Step 2: Separate the variables by writing the equation as: dy = dx + 2x y2 Step 3: Integrate both sides with respect to their respective variables: Z Z Z 1 dy = 1 dx + 2x dx y2 Step 4: Evaluate the integrals: 1 = x + x2 + C (where C is the constant of integration) y 1 = − = x2 + x + C y = −1 = y(x2 + x + C) 1 =y=− 2 x +x+C =− Step 5: Apply the initial condition y(0) = 1 to solve for the constant C: 1=− 02 1 +0+C 1 C C = −1 1=− Step 6: Substitute the value of C back into the solution found in Step 4 to get the final solution: 1 y=− 2 x +x−1 1 . Therefore, the solution to the initial value problem is y = − x2 +x−1 Question 28 Question Find the solution to the initial value problem: dy = 2x − 1, dx 27 y(0) = 3 Solution Step 1: Find the general solution by integrating both sides of the differential equation. Z Z dy dx = (2x − 1) dx dx Step 2: Integrate to find the general solution. Z y = (2x − 1) dx Step 3: Integrate to find the general solution. y = x2 − x + C Step 4: Use the initial condition y(0) = 3 to find the value of the constant C. 3 = 02 − 0 + C =⇒ C = 3 Step 5: Substitute the value of C back into the general solution to find the particular solution. y = x2 − x + 3 Therefore, the solution to the initial value problem is y = x2 − x + 3 with the initial condition y(0) = 3. Question 29 Question Consider the initial value problem: dy = 2x + y, dx y(0) = 1 Find the particular solution to the initial value problem. Solution Step 1: First, we rewrite the initial R value problem using an integrating factor. The integrating factor is µ(x) = e 1 dx = ex . Multiply the integrating factor by the original differential equation to get: ex dy − 2xex = yex dx Step 2: Notice that the left side is the derivative of a product. We can rewrite the equation as: d (yex ) = 0 dx 28 Step 3: Integrate both sides with respect to x: Z Z d x (ye ) dx = 0 dx dx yex = C where C is the constant of integration. Step 4: To find the particular solution, use the initial condition y(0) = 1: 1 · e0 = C C=1 Step 5: Therefore, the particular solution to the initial value problem is yex = 1, or y = e−x . Question 30 Question Solve the initial value problem: 2x + y dy = , dx x y(1) = 3 Solution Step 1: Rewrite the differential equation in standard form by multiplying through by dx: x dx = dy 2x + y Step 2: Use the initial condition y(1) = 3 to find the constant of integration: Z x Z y x dx = dy 1 3 2x + y Step 3: Evaluate the integrals: y Z x−1= 3 x dy 2x + y Step 4: Perform the integration on the right-hand side: Z y x y dy x − 1 = [x ln |2x + y|]3 − 3 2x + y Step 5: Evaluate the integral on the right-hand side by substitution: y x − 1 = y ln |2y + y| − 3 ln |2 · 3 + 3| − [x ln |2x + y|]3 Step 6: Simplify the expression using the initial condition: x − 1 = y ln |3y| − 3 ln 9 − (3 ln 9) − ln 3 + (1 ln 5) Step 7: Use the initial condition to find the final solution: x = y ln |3y| − ln 3 29 Question 31 Question Consider the initial value problem given by: dy x2 + y 2 = , dx xy y(1) = 1 Use the method of separation of variables to find the particular solution to the initial value problem. Solution To solve the initial value problem using separation of variables, we will need to dy express the given differential equation in the form of dx = f (x)g(y). Step 1: Rewrite the given differential equation to separate variables. x2 + y 2 x y dy = = + dx xy y x Step 2: Separate variables by multiplying through by y and dx: y dy = x dx + y dx Step 3: Integrate both sides: Z Z Z y dy = x dx + y dx 1 2 1 y = x2 + xy + C1 2 2 where C1 is the constant of integration. Step 4: Apply the initial condition y(1) = 1 to find the value of C1 : 1 2 1 (1) = (1)2 + (1)(1) + C1 2 2 1 1 = + 1 + C1 2 2 3 C1 = − 2 3 Step 5: Substitute C1 = − 2 back into the general solution: 1 2 1 3 y = x2 + xy − 2 2 2 y 2 = x2 + 2xy − 3 Therefore, the particular solution to the initial value problem is y 2 = x2 + 2xy − 3. 30 Question 32 Question Consider the initial value problem: y ′ = 2y − 3, y(0) = 4 Find the solution to the differential equation and determine the value of y when x = 3. Solution Step 1: First, let’s find the general solution to the differential equation y ′ = 2y − 3. The differential equation is linear, so we will first find the integrating R factor µ(x) = e 2 dx = e2x . Multiplying both sides of the differential equation by µ(x), we get: e2x y ′ − 2e2x y = −3e2x This can be rewritten as: d 2x (e y) = −3e2x dx Integrating both sides with respect to x, we get: Z 3 e2x y = − 3e2x dx = − e2x + C 2 Solving for y, we have: 3 y = − + Ce−2x 2 Step 2: Next, plug in the initial condition y(0) = 4 to solve for C. 3 4 = − + Ce0 2 3 11 = 2 2 Step 3: Substitute the value of C back into the general solution to get the particular solution. 3 11 y = − + e−2x 2 2 Step 4: Finally, find the value of y when x = 3. C =4+ 3 11 3 11 y(3) = − + e−6 = − + 6 2 2 2 2e 31 Question 33 Question Solve the initial value problem: 2y ′ + y = 4e−2x , y(0) = 3 Solution Step 1: First, we need to solve the homogeneous equation 2y ′ + y = 0 by finding the integrating factor µ(x). The integrating factor µ(x) is given by R 1 1 µ(x) = e 2 dx = e 2 x . Step 2: Multiply both sides of the homogeneous equation by the integrating factor µ(x): 1 1 e 2 x · (2y ′ + y) = e 2 x · 0 1 1 2e 2 x y ′ + e 2 x y = 0 Step 3: Recognize that the left-hand side of the above equation is the result of the product rule for differentiation. Therefore, we can rewrite it as follows: 1 d (2e 2 x y) = 0 dx Step 4: Integrate both sides of the above equation: Z Z 1 d x 2 (2e y)dx = 0dx dx 1 2e 2 x y = C Step 5: Solve for y in the above equation to find the general solution to the homogeneous equation: 1 y = Ce− 2 x Step 6: Next, we find a particular solution to the non-homogeneous equation by assuming yp = Ae−2x , where A is a constant coefficient to be determined. Step 7: Differentiate yp and substitute it into the non-homogeneous equation to solve for A: yp′ = −2Ae−2x 2(−2Ae−2x ) + Ae−2x = 4e−2x Step 8: Solve for A by simplifying the above equation: −4Ae−2x + Ae−2x = 4e−2x −3Ae−2x = 4e−2x 4 A=− 3 32 Step 9: Write the particular solution yp with the determined value of A: 4 yp = − e−2x 3 Step 10: The general solution to the non-homogeneous equation is the sum of the general solution to the homogeneous equation and the particular solution: 1 4 y = Ce− 2 x − e−2x 3 Step 11: Apply the initial condition y(0) = 3 to find the value of C: 4 3 = Ce0 − e0 3 3=C− C= 4 3 13 3 Step 12: The solution to the initial value problem 2y ′ + y = 4e−2x , y(0) = 3, is: y= 13 − 1 x 4 −2x e 2 − e 3 3 Question 34 Question Solve the initial value problem: dy = 2x(3y − 1) dx with y(0) = 2 Solution Step 1: Rewrite the differential equation in a more manageable form by separating variables. dy = 2xdx 3y − 1 Step 2: Integrate both sides with respect to their respective variables. Z Z dy = 2xdx 3y − 1 Step 3: Solve the integrals separately. Z Z dy = 2xdx 3y − 1 33 Step 4: For the left integral, use substitution by letting u = 3y − 1 and du = 3dy. Z Z 1 du = 2xdx 3 u Step 5: Integrate both sides. 1 ln |u| = x2 + C1 3 Step 6: Substitute back u = 3y − 1. 1 ln |3y − 1| = x2 + C1 3 Step 7: Solve for y. ln |3y − 1| = 3x2 + 3C1 Step 8: Move from logarithmic form to exponential form. 3y − 1 = e3x 2 +3C1 Step 9: Simplify the expression. 3y = e3x 2 +3C1 +1 Step 10: Solve for y by dividing by 3. e3x 2 +3C1 +1 3 Step 11: Use the initial condition y(0) = 2 to solve for the constant C1 . y= e0+3C1 + 1 3 Step 12: Simplify and solve for C1 . 2= 6 = e3C1 + 1 e3C1 = 5 3C1 = ln(5) ln(5) 3 Step 13: Substitute the value of C1 back into the solution obtained. C1 = y= e3x 2 +ln(5) +1 3 2 5e3x + 1 y= 3 2 Therefore, the solution to the initial value problem is y = 34 5e3x +1 . 3 Question 35 Question Consider the initial value problem: dy 2x + 3y = , dx 3x + 4y y(1) = 2 Find the particular solution to the initial value problem. Solution Step 1: Rewrite the initial value problem using separation of variables: dy 2x + 3y = dx 3x + 4y (3x + 4y)dy = (2x + 3y)dx Step 2: Integrate both sides: Z Z (3x + 4y)dy = (2x + 3y)dx 3xy + 2y 2 = x2 + 3xy + C 2y 2 = x2 + C Step 3: Use the initial condition y(1) = 2 to solve for the constant C: 2(2)2 = (1)2 + C 8=1+C C=7 Step 4: Substitute C = 7 back into the equation from Step 2: 2y 2 = x2 + 7 Step 5: Using the initial condition y(1) = 2, solve for x: 2(2)2 = (1)2 + 7 8=8 Step 6: Therefore, the particular solution to the initial value problem is 2y 2 = x2 + 7. 35 Step 3: Integrate both sides: Z Z 1 dy = dt y(y − 4)  Z  Z 1 1 dy = dt − + 4y 4(y − 4) 1 1 − ln |y| + ln |y − 4| = t + C 4 4 y−4 = 4t + C ln y y−4 = ke4t y where k = eC . Step 4: Apply initial condition: Substitute y = 3 and t = 0 into the solution to find k: 3−4 1 =k·1 ⇒ k =− 3 3 Step 5: Final solution: Thus, the solution to the initial value problem is: y−4 1 = − e4t y 3 y= 4 1 + 3e4t Question 2 Question Consider the initial value problem: dy = 2y − x, y(0) = 1 dx Find the particular solution to the initial value problem. Solution Step 1: First, we need to solve the differential equation. dy = 2y − x Given: dx Separating variables, we get: dy = dx 2y − x Integrating both sides gives: 2 Z 1 dy = 2y − x Z dx Step 2: Continuing from the previous step, we solve the integrals: Let u = 2y − x, then du = 2dy Therefore, the integral on the left side becomes: Z 1 1 1 du = ln |u| + C1 2 u 2 And the integral on the right side is simply: Z dx = x + C2 Step 3: Next, we substitute back for u and simplify: 1 ln |2y − x| = x + C 2 Where C = C2 − C1 is a constant of integration. Step 4: Now, we apply the initial condition y(0) = 1 to find the particular solution. When x = 0 and y = 1, we have: 1 ln |2(1) − 0| = 0 + C 2 1 ln 2 = C 2 So, the particular solution is: 1 1 ln |2y − x| = x + ln 2 2 2 Question 3 Question Consider the initial value problem: dy = 2y − x2 , dx y(0) = 1. Find the particular solution of the differential equation that satisfies the initial condition. 3 Solution Step 1: First, let’s solve the differential equation by separating variables. We have: dy = 2y − x2 . dx Step 2: Rearranging the terms, we get: dy = dx. 2y − x2 Step 3: Now, integrate both sides with respect to y and x separately. Z Z 1 dy = dx. 2y − x2 Step 4: The integral on the left side can be simplified by using substitution. Let u = 2y − x2 , then du = 2 dy. Z Z 1 1 du = dx. 2 u Step 5: Integrating, we have: 1 ln |u| = x + C, 2 where C is the constant of integration. Step 6: Substitute back u = 2y − x2 into the equation: 1 ln |2y − x2 | = x + C. 2 Step 7: Since y(0) = 1, substitute x = 0 and y = 1 into the equation to solve for C: 1 ln |2(1) − 02 | = 0 + C. 2 1 ln |2| = C. 2 C = ln |2|1/2 . Step 8: Finally, substitute the value of C back into the equation to find the particular solution: 1 ln |2y − x2 | = x + ln |2|1/2 . 2 Therefore, the particular solution of the differential equation that satisfies the initial condition y(0) = 1 is 21 ln |2y − x2 | = x + ln |2|1/2 . 4 Question 4 Question x Solve the initial value problem: y ′ = 12 y + 3e 2 , y(0) = 1. Solution Step 1: First, we find the integrating factorR for the given differential equation. x 1 The integrating factor is given by µ(x) = e 2 dx = e 2 . Step 2: Multiply both sides of the differential equation by the integrating factor: x 1 x e 2 y′ − e 2 y = 3 2 Step 3: Notice that the left side is the result of applying the product rule to x e 2 y. So we can rewrite the equation as: d x (e 2 y) = 3 dx Step 4: Integrate both sides with respect to x: Z Z d x (e 2 y)dx = 3dx dx x e 2 y = 3x + C Step 5: Apply the initial condition y(0) = 1 to find the value of C: e0 · 1 = 3 · 0 + C C=1 Step 6: Substitute the value of C back into the equation found in Step 4: x e 2 y = 3x + 1 Step 7: Solve for y: x x y = 3xe− 2 + e− 2 x x Therefore, the solution to the initial value problem is y = 3xe− 2 + e− 2 . Question 5 Question Solve the initial value problem: dy dx + 2xy = x, y(0) = 1. 5 Solution To solve this differential equation, we will use an integrating factor method. Step 1: Identify the integrating factor. The integrating factor µ(x) is given by: R µ(x) = e 2x dx Step 2: Find the integrating factor. Z 2x dx = x2 Hence, the integrating factor is: µ(x) = ex 2 Step 3: Multiply the differential equation by the integrating factor. The differential equation becomes: 2 ex 2 2 dy + 2xex y = xex dx Step 4: Rewrite the left-hand side as the derivative of a product. Using the product rule, we have: 2 d  x2  e y = xex dx Step 5: Integrate both sides. Integrating both sides gives: Z 2 2 ex y = xex dx + C 1 x2 e +C 2 Step 6: Solve for y. Solving for y, we get: 2 ex y = y= 2 1 + Ce−x 2 Step 7: Apply the initial condition. Using the initial condition y(0) = 1, we find: 1 1= +C 2 1 C= 2 Step 8: Final solution. Therefore, the solution to the initial value problem is: 2 1 1 y = + e−x 2 2 6 Question 6 Question Consider the initial value problem y ′ − 2y = e2x , where y(0) = 1. Find the particular solution to the differential equation that satisfies the initial condition. Solution Step 1: First, we find the integrating factor I(x). Let y ′ − 2y = e2x . The integrating factor is given by I(x) = e R −2dx = e−2x . Step 2: Multiply both sides of the differential equation by the integrating factor and simplify. Multiply e−2x to both sides of the equation to get e−2x (y ′ − 2y) = e−2x e2x . This simplifies to d −2x (e y) = 1. dx Step 3: Integrate both sides of the equation. Integrating both sides with respect to x, we get Z Z d −2x (e y)dx = 1dx. dx So, we have e−2x y = x + C, where C is the constant of integration. Step 4: Apply the initial condition y(0) = 1 to solve for the constant C. Substitute x = 0 and y = 1 into the equation above: e0 · 1 = 0 + C. Therefore, C = 1. Step 5: Finally, find the particular solution. Substitute the value of C back into the equation above: e−2x y = x + 1. Hence, the particular solution to the initial value problem is given by y = e2x (x + 1). Question 7 Question Solve the initial value problem: dy = −2x(y + 1), dx 7 y(0) = 2 Solution Step 1: We start by separating the variables in the given differential equation: dy = −2xdx y+1 Step 2: Next, we integrate both sides of the equation: Z Z 1 dy = −2xdx y+1 Step 3: Integrating the left side with respect to y and the right side with respect to x, we get: ln |y + 1| = −x2 + C where C is the constant of integration. Step 4: Exponentiating both sides of the equation yields: |y + 1| = e−x 2 +C = eC e−x 2 Step 5: Since y(0) = 2, we substitute x = 0 and y = 2 into the equation to find C: |2 + 1| = eC ⇒ 3 = eC ⇒ C = ln 3 Step 6: Therefore, the solution to the initial value problem is: 2 y = e−x − 1 Question 8 Question Consider the initial value problem: dy = 2x2 − 4x + 5, dx y(0) = 3. Find the solution to the initial value problem. Solution Step 1: Find the general solution dy = 2x2 − 4x + 5. dx Z Z dy = (2x2 − 4x + 5)dx. y= 2 3 x − 2x2 + 5x + C. 3 8 where C is the constant of integration. Step 2: Use the initial condition to find the particular solution y(0) = 3, 2 3 (0) − 2(0)2 + 5(0) + C = 3, 3 C = 3. Therefore, the solution to the initial value problem is: y= 2 3 x − 2x2 + 5x + 3. 3 Question 9 Question Solve the initial value problem: y ′ = sin(2x) − 2y, y(0) = 1 Solution Step 1: First, we find the general solution to the homogeneous differential equation y ′ = −2y by separating variables: dy dy = −2y ⇒ = −2dx ⇒ ln |y| = −2x + C1 dx y where C1 is the constant of integration. Exponentiating both sides gives |y| = eC1 e−2x . Since eC1 is just another constant, we can simplify this to |y| = Ce−2x , where C is a constant. Step 2: Now, we will find a particular solution to the non-homogeneous part of the equation. Let yp be a particular solution to the equation y ′ = sin(2x). We can guess yp = A sin(2x) + B cos(2x) and find A and B by substitution: yp′ = 2A cos(2x) − 2B sin(2x) = sin(2x) Comparing coefficients, we get 2A = 0 and −2B = 1, which gives A = 0 and B = − 21 . Therefore, yp = − 12 cos(2x). Step 3: The general solution to the non-homogeneous differential equation is the sum of the particular solution and the general solution to the homogeneous equation, so y = Ce−2x − 12 cos(2x). Step 4: Using the initial condition y(0) = 1, we find: 1 = Ce0 − 1 3 1 cos(0) ⇒ 1 = C − ⇒ C = 2 2 2 Therefore, the solution to the initial value problem is y = 32 e−2x − 12 cos(2x). 9 Question 10 Question Consider the initial value problem: dy 3x2 + y = , dx 2x y(1) = 4 Find the particular solution to the initial value problem. Solution Step 1: Rewrite the given differential equation in a more convenient form. dy 3x2 + y dy = =⇒ 2x = 3x2 + y dx 2x dx Step 2: Recognize that the differential equation is linear and in standard form y ′ + P (x)y = Q(x), where P (x) = x2 and Q(x) = 3x2 . R P (x)dx Step 3: Find the integrating factor µ(x) = e . µ(x) = e R 2 x dx 2 = e2 ln |x| = eln x = x2 Step 4: Multiply the integrating factor by both sides of the differential equation. dy x2 · − 2xy = 3x3 dx Step 5: Rewrite the equation as the derivative of a product. d 2 (x y) = 3x3 dx Step 6: Integrate both sides of the equation with respect to x. Z Z d(x2 y) = 3x3 dx Step 7: Solve the integrals. x2 y = 3 4 x +C 4 Step 8: Apply the initial condition y(1) = 4 to find the value of the constant C. 3 4 3 16 3 13 (1) + C =⇒ 4 = + C =⇒ C = − = 4 4 4 4 4 Step 9: Write the particular solution. (1)2 · 4 = x2 y = 3 4 13 3 13 x + =⇒ y = x2 + 2 4 4 4 4x Therefore, the particular solution to the initial value problem is y = 34 x2 + 13 4x2 . 10 Question 11 Question Consider the initial value problem: dy 2x − y = , dx 2y − x y(0) = 1 Solve the initial value problem. Solution To solve the initial value problem, we will first rewrite the given differential equation in a more familiar form by letting u = xy . Step 1: Rewrite the differential equation using u = y x Let u = xy . dy = x du Then, we have y = ux and dx dx + u. y Substitute u = x into the given differential equation to obtain: x du 2x − ux +u= dx 2ux − x Step 2: Solve for u Solving the above equation for u gives: x x x du 2x − ux +u= dx 2ux − x 2x(1 − u) du +u= dx x(2u − 1) x 2(1 − u) du +u= dx 2u − 1 x du 2(1 − u) = −u dx 2u − 1 2 − 2u − 2u(2u − 1) du = dx 2u − 1 x du 2 − 2u − 4u2 + 2u = dx 2u − 1 du 2 − 4u2 = dx 2u − 1 Separate variables, x and u, to solve for u by integrating: x 1 (2u − 1)du = (2 − 4u2 ) dx x 11 Question 12 Question Solve the initial value problem: y′ = 1 , x2 + y y(1) = −1. Solution Step 1: Rewrite the differential equation in standard form. 1 dy = 2 dx x +y Step 2: Separate variables to solve the differential equation. (x2 + y) dy = dx Step 3: Integrate both sides of the equation. Z Z 2 (x + y) dy = dx 1 x2 y + y 2 = x + C 2 Step 4: Use the initial condition y(1) = −1 to find the value of C. 1 12 (−1) + (−1)2 = 1 + C 2 −1 − 1 =1+C 2 3 =1+C 2 5 C=− 2 Step 5: Substitute the value of C back into the equation found in Step 3. − 1 5 x2 y + y 2 = x − 2 2 Step 6: Now, we need to solve for y. This is a nonlinear equation, and solving for y explicitly may be difficult. Therefore, we have found the general solution implicitly. 12 Question 13 Question Find the particular solution to the initial value problem: y ′ = 2x + 3e2y , y(0) = 1. Solution Step 1: First, separate the variables in the given differential equation: dy = 2x + 3e2y dx Step 2: Next, integrate both sides with respect to x: Z Z Z dy = 2xdx + 3e2y dx Step 3: Integrate each term separately: y = x2 + C1 + 3 Z e2y dx Step 4: To integrate e2y with respect to y, let u = 2y so du = 2dy: Z 3 u e du y = x2 + C1 + 2 Step 5: Evaluate the integral: 3 y = x2 + C1 + e2y + C2 2 Step 6: Simplify the equation: 3 y − e2y = x2 + C 2 Step 7: Use the initial condition y(0) = 1 to find the value of the constant C: 3 1 − e2 = 0 + C 2 Step 8: Solve for C: 3 2 e −1 2 Step 9: Substitute the value of C back into the equation to find the particular solution: 3 3 y − e2y = x2 + e2 − 1 2 2 Therefore, the particular solution to the initial value problem is y = x2 + 3 2 3 2y 2e − 1 + 2e . C= 13 Question 14 Question Solve the initial value problem: dy = t2 − 2y, dt y(0) = 1 Solution Step 1: First, let’s find the general solution to the differential equation dy dt = dy 2 2 t − 2y. We can rewrite this equation as dt + 2y = t . Step 2: This is a first-order linear differential equation, so we can use the R integrating factor method. The integrating factor is e 2 dt = e2t . Step 3: Multiply the integrating factor by the equation to get: e2t dy + 2e2t y = e2t t2 dt Step 4: Rewrite the left side as a derivative of a product: d 2t (e y) = e2t t2 dt Step 5: Integrate both sides with respect to t: Z Z d 2t (e y) dt = e2t t2 dt dt Step 6: Integrating gives: e2t y = Z t2 e2t dt Step 7: Now, we need to solve the integral on the right side. This requires integrating by parts one or more times. Step 8: After integrating by parts, we find: Z Z 1 2 2t 1 2 2t t e dt = t e − e2t dt 2 2 Step 9: Continuing to solve the integral, we have: Z 1 e2t dt = e2t + C 2 Step 10: Substituting this back in, we have: e2t y = 1 2 2t 1 2t C t e − e − 2 4 2 14 Step 11: Solve for y by dividing by e2t : y= C 1 2 1 t − − 2t 2 4 2e Step 12: Use the initial condition y(0) = 1 to find the value of C. Substitute t = 0 and y = 1 into the equation. Step 13: Solving for C gives C = −1/4. Step 14: Substitute C back into the equation for y to obtain the final solution. Therefore, the solution to the initial value problem is: y= 1 2 1 1 t − − 2t 2 4 2e Question 15 Question Find the particular solution to the initial value problem: dy = 3x2 + 4x, dx y(0) = 5. Solution Step 1: First, we solve the differential equation to find the general solution. dy = 3x2 + 4x dx Integrating both sides with respect to x, we get: Z Z dy = (3x2 + 4x)dx y = x3 + 2x2 + C where C is a constant. Step 2: Next, we apply the initial condition y(0) = 5 to find the particular solution. 5 = 03 + 2(0)2 + C C=5 Step 3: Therefore, the particular solution to the initial value problem is: y = x3 + 2x2 + 5 15 Question 16 Question Solve the initial value problem: dy = 2xy, dx y(0) = −1. Solution Step 1: Write the initial value problem as a differential equation. dy = 2xy, y(0) = −1. dx Step 2: Separate variables to solve the differential equation. dy = 2xdx. y Step 3: Integrate both sides. Z 1 dy = y Z 2xdx. Step 4: Solve the integrals. ln |y| = x2 + C, where C is the constant of integration. Step 5: Solve for y. 2 2 y = ex +C = Cex . Step 6: Use the initial condition y(0) = −1 to find the value of C. −1 = C. Step 7: Substitute C = −1 back into the equation for y. 2 y = −ex . 2 Therefore, the solution to the initial value problem is y = −ex . Question 17 Question Solve the initial value problem: dy 1 = y dx 2 with y(0) = 3 16 Solution Step 1: First, we rearrange the differential equation to separate variables: dy 1 = dx y 2 Step 2: Next, we integrate both sides: Z Z 1 1 dy = dx y 2 Step 3: Integrating, we get: ln |y| = 1 x+C 2 Step 4: Using the initial condition y(0) = 3, we find the value of the constant C: ln |3| = 0 + C =⇒ C = ln(3) Step 5: Substituting this value back into the equation, we have: ln |y| = 1 x + ln(3) 2 Step 6: Exponentiating both sides to eliminate the natural logarithm: 1 1 y = e 2 x+ln(3) = 3e 2 x 1 Therefore, the solution to the initial value problem is y = 3e 2 x . Question 18 Question Consider the initial value problem: dy = e2x sin(3x), dx y(0) = 1 Find the particular solution to the initial value problem. Solution Step 1: First, we need to solve the differential equation grating both sides with respect to x gives: Z Z dy dx = e2x sin(3x) dx dx 17 dy dx = e2x sin(3x). Inte- 1 2x e (3 sin(3x) − cos(3x)) + C 13 Step 2: Next, we use the initial condition y(0) = 1 to find the value of C. Substitute x = 0 and y = 1 into the equation: y=− 1=− 1 (3 sin(0) − cos(0)) + C 13 1 (−1) + C 13 1 12 1= +C ⇒C = 13 13 12 Step 3: Finally, substitute C = 13 back into the general solution to find the particular solution: 1=− y=− 12 1 2x e (3 sin(3x) − cos(3x)) + 13 13 1 2x Therefore, the particular solution to the initial value problem is y = − 13 e (3 sin(3x)− 12 cos(3x)) + 13 . Question 19 Question Solve the initial value problem: dy dx = e−2y , y(0) = 1. Solution To solve this initial value problem, we will separate variables and then integrate both sides. Step 1: Separate variables by writing the differential equation as follows: dy = e−2y dx e2y dy = dx Step 2: Integrate both sides: Z e2y dy = Z dx 1 2y e =x+C 2 where C is the constant of integration. Step 3: Solve for y using the initial condition y(0) = 1: 1 2·1 e =0+C 2 18 1 2 e =C 2 Step 4: Substitute back into the equation to find the solution: 1 2y 1 e = x + e2 2 2 1 e2y = 2(x + e2 ) 2
2y = ln 2x + e2 y=
1 ln 2x + e2 2 Therefore, the solution to the initial value problem is y = 1 2
ln 2x + e2 . Question 20 Question Consider the initial value problem: y′ = 2y + sin(t), t y(1) = 1. Find the particular solution to the initial value problem. Solution Step 1: First, we rewrite the given initial value problem in standard form by separating variables: dy 2y = + sin(t). dt t Step 2: Next, we isolate y terms and t terms: dt dy = + sin(t)dt. 2y t Step 3: We integrate both sides: Z Z Z 1 1 dy = dt + sin(t) dt. 2y t Step 4: Solving the integrals gives us: 1 ln |y| = ln |t| − cos(t) + C, 2 where C is the constant of integration. 19 Step 5: Simplifying and solving for y, we have: ln |y| = 2 ln |t| − 2 cos(t) + C ′ . Here, C ′ = 2C. Step 6: Exponentiating both sides, we obtain: |y| = e2 ln |t| · e−2 cos(t)+C ′ ′ |y| = t2 e−2 cos(t)+C . Step 7: Applying the initial condition y(1) = 1, we get: ′ |1| = 12 e−2 cos(1)+C , which gives us C ′ = 0. Step 8: Therefore, the particular solution to the initial value problem is: y = t2 e−2 cos(t) . Question 21 Question Consider the initial value problem dy = 2y − x, dx y(0) = 1. Find the particular solution y(x) satisfying the initial condition. Solution Step 1: First, let’s solve the differential equation using the method of integrating R factor. The integrating factor is given by µ(x) = e 2 dx = e2x . Multiplying both sides of the differential equation by the integrating factor, we have e2x dy = e2x (2y − x). dx Step 2: Notice that the left side can be written as the derivative of the product e2x y. Therefore, we have d 2x (e y) = e2x (2y − x). dx Step 3: Integrating both sides with respect to x, we get Z 2x e y = e2x (2y − x) dx. 20 Step 4: Solving the integral on the right side, we have Z Z 2x 2x e y = 2 e y dx − xe2x dx. R Step 5: Integrating xe2x dx by parts, let u = x and dv = e2x dx, then du = dx and v = 21 e2x . Hence, we have 1 1 e2x y = 2e2x y − xe2x − e2x + C, 2 4 where C is the constant of integration. Step 6: Simplifying the equation, we get 1 1 e2x y − 2e2x y = − xe2x − e2x + C. 2 4 Step 7: Continuing to simplify, we find 1 1 −y = − x − + Ce−2x . 2 4 Step 8: Finally, applying the initial condition y(0) = 1, we get −1 = 0 − 1 +C 4 3 C=− . 4 ⇒ Step 9: Therefore, the particular solution to the initial value problem is 1 1 3 y(x) = − x − − e−2x . 2 4 4 Question 22 Question Solve the initial value problem: dy = 2x2 − 3x, dx y(1) = 4 Solution Step 1: Find the general solution by solving the differential equation. dy = 2x2 − 3x dx Integrating both sides with respect to x, we get: Z y = (2x2 − 3x)dx 21 2 3 3 2 x − x +C 3 2 where C is the constant of integration. Step 2: Use the initial condition y(1) = 4 to find the value of C. y= 4= 2 3 3 2 (1) − (1) + C 3 2 2 3 2 4= 3 2 4= 3 4 4= 6 3 +C 2 9 − +C 6 3 − +C 6 3 − +C 6 1 4= +C 6 1 C =4− 6 23 C= 6 Step 3: Substitute the value of C back into the general solution. 4= y= − 2 3 3 2 23 x − x + 3 2 6 Therefore, the solution to the initial value problem is: y= 2 3 3 2 23 x − x + 3 2 6 Question 23 Question Solve the initial value problem: dy = 3y − 2x, dx 22 y(0) = 4 Solution Step 1: First, we write the given differential equation in the standard form dy dx − 3y = −2x. R Step 2: Next, we find the integrating factor by computing e −3 dx . Integrating factor = e−3x Step 3: We multiply the integrating factor with both sides of the differential equation: dy − 3e−3x y = −2xe−3x e−3x dx Step 4: The left-hand side of the equation can be rewritten using the product d (ye−3x ). rule as dx d (ye−3x ) = −2xe−3x dx Step 5: Now, we integrate both sides with respect to x: Z Z d (ye−3x ) dx = −2xe−3x dx dx Step 6: Integrating both sides gives us: ye−3x = 2 2 −3x x e +C 9 where C is the constant of integration. Step 7: To find the value of C, we use the initial condition y(0) = 4. ⇒ 4=0+C C=4 Step 8: Substituting the value of C back into the equation, we have: ye−3x = 2 2 −3x x e +4 9 Step 9: Finally, solving for y gives: y= 2 2 x + 4e3x 9 Therefore, the solution to the initial value problem is y = 29 x2 + 4. Question 24 Question Consider the initial value problem: dy = x2 + y, dx y(0) = 2 Find the particular solution to the initial value problem. 23 Solution Step 1: First, let’s rewrite the given initial value problem into a more familiar form: dy − y = x2 , y(0) = 2 dx dy + Step 2: This is a first-order linear differential equation of the form dx 2 P (x)y = Q(x), where P (x) = −1 and Q(x) = x . Step 3: To solveR the differential equation, we Rmultiply both sides by the integrating factor, e P (x)dx , which in this case is e −1dx = e−x . This gives us: e−x dy − e−x y = e−x x2 dx Step 4: Notice that the left-hand side is equal to rule. Therefore, we rewrite the equation as: d −x ) dx (ye by the product d (ye−x ) = e−x x2 dx Step 5: Now we integrate both sides with respect to x to solve for y: Z Z d −x (ye )dx = e−x x2 dx dx Step 6: Integrating both sides gives us: ye−x = −e−x x2 − 2e−x + C where C is the constant of integration. Step 7: To solve for y, multiply through by ex and simplify: y = −x2 − 2 + Cex Step 8: Using the initial condition, y(0) = 2, we can find the value of the constant C: 2 = −02 − 2 + Ce0 2 = −2 + C ⇒ C = 4 Step 9: Therefore, the particular solution to the initial value problem is: y = −x2 − 2 + 4ex Question 25 Question Solve the initial value problem: dy = 2x + 3y, dx 24 y(0) = 1 Solution Step 1: First, let’s rewrite the given initial value problem as: dy − 3y = 2x dx with y(0) = 1 Step 2: This is a first-order linear differential equation. We can solve it by finding the integrating factor. The integrating factor I(x) is given by: I(x) = e R −3dx = e−3x Step 3: Now, multiply both sides of the differential equation by the integrating factor: dy − 3e−3x y = 2xe−3x e−3x dx Step 4: The left side can be written in the form of the product rule: d −3x (e y) = 2xe−3x dx Step 5: Integrate both sides with respect to x: Z Z −3x d(e y) = 2xe−3x dx Step 6: This gives us: Z 2 2 e−3x y = 2xe−3x dx = − xe−3x − e−3x + C 3 9 Step 7: Solving for y, we get: 2 2 y = − x − + Ce3x 3 9 Step 8: Using the initial condition y(0) = 1, we can find the value of the constant C: 2 11 1=− +C ⇒ C = 9 9 Step 9: Therefore, the solution to the initial value problem is: 2 2 11 y = − x − + e3x 3 9 9 Question 26 Question Solve the initial value problem: y′ = 2x + y , x + 2y 25 y(0) = 1 Solution To solve the initial value problem, we will use separation of variables. Let’s begin by rewriting the differential equation: y′ = 2x + y x + 2y Step 1: Write the equation in the form g(y) dy = f (x) dx: (x + 2y) dy = (2x + y) dx Step 2: Integrate both sides: Z Z (x + 2y) dy = (2x + y) dx Z Z x dy + Z 2y dy = Z 2x dx + y dx xy + y 2 = x2 + y + C Step 3: Use the initial condition y(0) = 1 to solve for the constant C: 0 ∗ 1 + 12 = 02 + 1 + C C=1 Step 4: Substitute C = 1 back into the equation: xy + y 2 = x2 + y + 1 Step 5: Since we cannot solve explicitly for y, we have found the implicit solution to the differential equation: xy + y 2 = x2 + y + 1 Question 27 Question Consider the initial value problem: dy = y 2 + 2x, dx y(0) = 1 Find the solution to the initial value problem. 26 Solution Step 1: Rewrite the initial value problem as a separable differential equation: dy = y 2 + 2x dx Step 2: Separate the variables by writing the equation as: dy = dx + 2x y2 Step 3: Integrate both sides with respect to their respective variables: Z Z Z 1 dy = 1 dx + 2x dx y2 Step 4: Evaluate the integrals: 1 = x + x2 + C (where C is the constant of integration) y 1 = − = x2 + x + C y = −1 = y(x2 + x + C) 1 =y=− 2 x +x+C =− Step 5: Apply the initial condition y(0) = 1 to solve for the constant C: 1=− 02 1 +0+C 1 C C = −1 1=− Step 6: Substitute the value of C back into the solution found in Step 4 to get the final solution: 1 y=− 2 x +x−1 1 . Therefore, the solution to the initial value problem is y = − x2 +x−1 Question 28 Question Find the solution to the initial value problem: dy = 2x − 1, dx 27 y(0) = 3 Solution Step 1: Find the general solution by integrating both sides of the differential equation. Z Z dy dx = (2x − 1) dx dx Step 2: Integrate to find the general solution. Z y = (2x − 1) dx Step 3: Integrate to find the general solution. y = x2 − x + C Step 4: Use the initial condition y(0) = 3 to find the value of the constant C. 3 = 02 − 0 + C =⇒ C = 3 Step 5: Substitute the value of C back into the general solution to find the particular solution. y = x2 − x + 3 Therefore, the solution to the initial value problem is y = x2 − x + 3 with the initial condition y(0) = 3. Question 29 Question Consider the initial value problem: dy = 2x + y, dx y(0) = 1 Find the particular solution to the initial value problem. Solution Step 1: First, we rewrite the initial R value problem using an integrating factor. The integrating factor is µ(x) = e 1 dx = ex . Multiply the integrating factor by the original differential equation to get: ex dy − 2xex = yex dx Step 2: Notice that the left side is the derivative of a product. We can rewrite the equation as: d (yex ) = 0 dx 28 Step 3: Integrate both sides with respect to x: Z Z d x (ye ) dx = 0 dx dx yex = C where C is the constant of integration. Step 4: To find the particular solution, use the initial condition y(0) = 1: 1 · e0 = C C=1 Step 5: Therefore, the particular solution to the initial value problem is yex = 1, or y = e−x . Question 30 Question Solve the initial value problem: 2x + y dy = , dx x y(1) = 3 Solution Step 1: Rewrite the differential equation in standard form by multiplying through by dx: x dx = dy 2x + y Step 2: Use the initial condition y(1) = 3 to find the constant of integration: Z x Z y x dx = dy 1 3 2x + y Step 3: Evaluate the integrals: y Z x−1= 3 x dy 2x + y Step 4: Perform the integration on the right-hand side: Z y x y dy x − 1 = [x ln |2x + y|]3 − 3 2x + y Step 5: Evaluate the integral on the right-hand side by substitution: y x − 1 = y ln |2y + y| − 3 ln |2 · 3 + 3| − [x ln |2x + y|]3 Step 6: Simplify the expression using the initial condition: x − 1 = y ln |3y| − 3 ln 9 − (3 ln 9) − ln 3 + (1 ln 5) Step 7: Use the initial condition to find the final solution: x = y ln |3y| − ln 3 29 Question 31 Question Consider the initial value problem given by: dy x2 + y 2 = , dx xy y(1) = 1 Use the method of separation of variables to find the particular solution to the initial value problem. Solution To solve the initial value problem using separation of variables, we will need to dy express the given differential equation in the form of dx = f (x)g(y). Step 1: Rewrite the given differential equation to separate variables. x2 + y 2 x y dy = = + dx xy y x Step 2: Separate variables by multiplying through by y and dx: y dy = x dx + y dx Step 3: Integrate both sides: Z Z Z y dy = x dx + y dx 1 2 1 y = x2 + xy + C1 2 2 where C1 is the constant of integration. Step 4: Apply the initial condition y(1) = 1 to find the value of C1 : 1 2 1 (1) = (1)2 + (1)(1) + C1 2 2 1 1 = + 1 + C1 2 2 3 C1 = − 2 3 Step 5: Substitute C1 = − 2 back into the general solution: 1 2 1 3 y = x2 + xy − 2 2 2 y 2 = x2 + 2xy − 3 Therefore, the particular solution to the initial value problem is y 2 = x2 + 2xy − 3. 30 Question 32 Question Consider the initial value problem: y ′ = 2y − 3, y(0) = 4 Find the solution to the differential equation and determine the value of y when x = 3. Solution Step 1: First, let’s find the general solution to the differential equation y ′ = 2y − 3. The differential equation is linear, so we will first find the integrating R factor µ(x) = e 2 dx = e2x . Multiplying both sides of the differential equation by µ(x), we get: e2x y ′ − 2e2x y = −3e2x This can be rewritten as: d 2x (e y) = −3e2x dx Integrating both sides with respect to x, we get: Z 3 e2x y = − 3e2x dx = − e2x + C 2 Solving for y, we have: 3 y = − + Ce−2x 2 Step 2: Next, plug in the initial condition y(0) = 4 to solve for C. 3 4 = − + Ce0 2 3 11 = 2 2 Step 3: Substitute the value of C back into the general solution to get the particular solution. 3 11 y = − + e−2x 2 2 Step 4: Finally, find the value of y when x = 3. C =4+ 3 11 3 11 y(3) = − + e−6 = − + 6 2 2 2 2e 31 Question 33 Question Solve the initial value problem: 2y ′ + y = 4e−2x , y(0) = 3 Solution Step 1: First, we need to solve the homogeneous equation 2y ′ + y = 0 by finding the integrating factor µ(x). The integrating factor µ(x) is given by R 1 1 µ(x) = e 2 dx = e 2 x . Step 2: Multiply both sides of the homogeneous equation by the integrating factor µ(x): 1 1 e 2 x · (2y ′ + y) = e 2 x · 0 1 1 2e 2 x y ′ + e 2 x y = 0 Step 3: Recognize that the left-hand side of the above equation is the result of the product rule for differentiation. Therefore, we can rewrite it as follows: 1 d (2e 2 x y) = 0 dx Step 4: Integrate both sides of the above equation: Z Z 1 d x 2 (2e y)dx = 0dx dx 1 2e 2 x y = C Step 5: Solve for y in the above equation to find the general solution to the homogeneous equation: 1 y = Ce− 2 x Step 6: Next, we find a particular solution to the non-homogeneous equation by assuming yp = Ae−2x , where A is a constant coefficient to be determined. Step 7: Differentiate yp and substitute it into the non-homogeneous equation to solve for A: yp′ = −2Ae−2x 2(−2Ae−2x ) + Ae−2x = 4e−2x Step 8: Solve for A by simplifying the above equation: −4Ae−2x + Ae−2x = 4e−2x −3Ae−2x = 4e−2x 4 A=− 3 32 Step 9: Write the particular solution yp with the determined value of A: 4 yp = − e−2x 3 Step 10: The general solution to the non-homogeneous equation is the sum of the general solution to the homogeneous equation and the particular solution: 1 4 y = Ce− 2 x − e−2x 3 Step 11: Apply the initial condition y(0) = 3 to find the value of C: 4 3 = Ce0 − e0 3 3=C− C= 4 3 13 3 Step 12: The solution to the initial value problem 2y ′ + y = 4e−2x , y(0) = 3, is: y= 13 − 1 x 4 −2x e 2 − e 3 3 Question 34 Question Solve the initial value problem: dy = 2x(3y − 1) dx with y(0) = 2 Solution Step 1: Rewrite the differential equation in a more manageable form by separating variables. dy = 2xdx 3y − 1 Step 2: Integrate both sides with respect to their respective variables. Z Z dy = 2xdx 3y − 1 Step 3: Solve the integrals separately. Z Z dy = 2xdx 3y − 1 33 Step 4: For the left integral, use substitution by letting u = 3y − 1 and du = 3dy. Z Z 1 du = 2xdx 3 u Step 5: Integrate both sides. 1 ln |u| = x2 + C1 3 Step 6: Substitute back u = 3y − 1. 1 ln |3y − 1| = x2 + C1 3 Step 7: Solve for y. ln |3y − 1| = 3x2 + 3C1 Step 8: Move from logarithmic form to exponential form. 3y − 1 = e3x 2 +3C1 Step 9: Simplify the expression. 3y = e3x 2 +3C1 +1 Step 10: Solve for y by dividing by 3. e3x 2 +3C1 +1 3 Step 11: Use the initial condition y(0) = 2 to solve for the constant C1 . y= e0+3C1 + 1 3 Step 12: Simplify and solve for C1 . 2= 6 = e3C1 + 1 e3C1 = 5 3C1 = ln(5) ln(5) 3 Step 13: Substitute the value of C1 back into the solution obtained. C1 = y= e3x 2 +ln(5) +1 3 2 5e3x + 1 y= 3 2 Therefore, the solution to the initial value problem is y = 34 5e3x +1 . 3 Question 35 Question Consider the initial value problem: dy 2x + 3y = , dx 3x + 4y y(1) = 2 Find the particular solution to the initial value problem. Solution Step 1: Rewrite the initial value problem using separation of variables: dy 2x + 3y = dx 3x + 4y (3x + 4y)dy = (2x + 3y)dx Step 2: Integrate both sides: Z Z (3x + 4y)dy = (2x + 3y)dx 3xy + 2y 2 = x2 + 3xy + C 2y 2 = x2 + C Step 3: Use the initial condition y(1) = 2 to solve for the constant C: 2(2)2 = (1)2 + C 8=1+C C=7 Step 4: Substitute C = 7 back into the equation from Step 2: 2y 2 = x2 + 7 Step 5: Using the initial condition y(1) = 2, solve for x: 2(2)2 = (1)2 + 7 8=8 Step 6: Therefore, the particular solution to the initial value problem is 2y 2 = x2 + 7. 35