Initial Value Problems Question Bank Set 3

MATH 334 - DIFFERENTIAL EQUATIONS - Initial value problems Question Bank - Set 3 Liberty University Question 1 Question Consider the initial value problem: dy = 2x − 3y, y(0) = 1 dx Determine the particular solution to the initial value problem. Solution To solve this initial value problem, we can use R the method of integrating factors. The integrating factor is given by µ(x) = e −3 dx = e−3x . Step 1: Multiply both sides of the differential equation by the integrating factor µ(x). dy e−3x − 3e−3x y = 2xe−3x dx Step 2: Recognize the left-hand side as the result of the product rule: d (ye−3x ) = 2xe−3x dx Step 3: Integrate both sides with respect to x: Z Z d (ye−3x ) dx = 2xe−3x dx dx Z ye−3x = 2xe−3x dx + C Step 4: Solve the integral on the right-hand side: 2 2 ye−3x = e−3x (− x − ) + C 3 9 2 2 y = − x − + Ce3x 3 9 Step 5: Use the initial condition y(0) = 1 to find the value of C: 2 1=− +C 9 C =1+ Step 6: Substitute C = 11 9 11 2 = 9 9 back into the particular solution: 2 2 11 y = − x − + e3x 3 9 9 2 9 Therefore, the particular solution to the initial value problem is y = − 32 x − 3x + 11 9 e . Question 2 Question Solve the initial value problem: y ′ + 2y = 4e−2x , y(0) = 3 Solution Step 1: First, we find the integrating factor: Let P (x) = 2, then the integrating factor is given by: R R µ(x) = e P (x) dx = e 2 dx = e2x Step 2: Next, we multiply the given equation by the integrating factor µ(x): e2x y ′ + 2e2x y = 4e2x e−2x e2x y ′ + 2e2x y = 4 Step 3: Rewrite the left side of the equation as the derivative of the product y(x)e2x : (e2x y)′ = 4 Step 4: Integrate both sides with respect to x: Z Z 2x ′ (e y) dx = 4 dx 2 e2x y = 4x + C Step 5: Apply the initial condition y(0) = 3 to find the value of the constant C: e2·0 · 3 = 4 · 0 + C 3=C Step 6: Write the final solution by substituting the value of C back into the equation: e2x y = 4x + 3 y= 4x + 3 4x + 3 = 2x e e2x Question 3 Question Solve the initial value problem: y ′ = 2y − 4 with the initial condition y(0) = 3. Solution Step 1: First, we solve the differential equation y ′ = 2y − 4. This is a firstorder linear differential equation. To solve it, we rewrite it in the standard form y ′ − 2y = −4. Step 2: To solve the homogeneous part of the R equation, we find the integrating factor. The integrating factor is given by e −2 dx = e−2x . Step 3: Multiply the integrating factor by the given equation to get e−2x y ′ − 2e−2x y = −4e−2x . d Step 4: Rewriting the left side as the derivative of e−2x y, we have dx (e−2x y) = −2x −4e . Step 5: Integrating both sides with respect to x gives e−2x y = 2e−2x + C, where C is the constant of integration. Step 6: Solving for y, we have y = 2 + Ce2x . Step 7: To find the particular solution, we use the initial condition y(0) = 3. Substitute x = 0 and y = 3 into the equation y = 2 + Ce2x to find the value of C. Step 8: We have 3 = 2 + C, which implies C = 1. Step 9: Substituting C = 1 back into the equation y = 2 + Ce2x gives the particular solution y = 2 + e2x . Therefore, the solution to the initial value problem y ′ = 2y − 4 with y(0) = 3 is y = 2 + e2x . 3 Question 4 Question Solve the initial value problem: y ′ + 2xy = 2x, y(0) = 1 Solution Step 1: First, we find the integrating factor by calculating e Z 2xdx = x2 R 2xdx . 2 Therefore, the integrating factor is ex . Step 2: Multiply both sides of the differential equation by the integrating factor and rewrite the left-hand side as a product rule: 2 2 ex y ′ + 2xex y = 2xex 2 Step 3: Recognize that the left-hand side is now the derivative of a product: 2 (ex y)′ = 2xex 2 Step 4: Integrate both sides with respect to x: Z Z 2 x2 ′ (e y) dx = 2xex dx Z x2 e y= 2 2xex dx + C Step 5: Solve the integral on the right-hand side: 2 2 ex y = ex + C 2 Step 6: Solve for y by dividing by ex : y = 1 + Ce−x 2 Step 7: Use the initial condition y(0) = 1 to find the value of the constant C: 1 = 1 + C · e0 C=0 Step 8: Substitute the value of C back into the equation to obtain the particular solution: y=1 Therefore, the solution to the initial value problem is y = 1. 4 Question 5 Question Solve the initial value problem: dy = 3y − 6x, dx y(0) = 4 Solution Step 1: First, we identify the differential equation as a first-order linear equation and rewrite it in standard form dy/dx + P (x)y = Q(x), where P (x) and Q(x) are functions of x. dy = 3y − 6x dx dy − 3y = −6x dx Step 2: Next, we identify P (x) = −3 and Q(x) = −6x. We now find the R integrating factor I(x) = e P (x)dx . I(x) = e R −3dx = e−3x Step 3: We multiply the differential equation by the integrating factor I(x). e−3x dy − 3e−3x y = −6xe−3x dx d −3x (e y) = −6xe−3x dx Step 4: Integrate both sides of the equation with respect to x. Z Z d −3x (e y)dx = −6xe−3x dx dx Z −3x e y = −6xe−3x dx Step 5: Solve the integral on the right side. Z −3x −3x e y = 3xe − 3e−3x dx e−3x y = 3xe−3x + C Step 6: Multiply through by e3x . y = 3x + Ce3x 5 Step 7: Apply the initial condition y(0) = 4 to solve for C. 4 = 3(0) + C(1) C=4 Step 8: Substitute C = 4 back into the general solution. y = 3x + 4e3x Therefore, the solution to the initial value problem is y = 3x + 4e3x . Question 6 Question Consider the initial value problem: dy = y2 , dx y(0) = 1. Find the particular solution to the differential equation that satisfies the initial condition. Solution Step 1: Separate the variables: 1 dy =1 y 2 dx Step 2: Integrate both sides with respect to x: Z Z 1 dy dx = 1 dx y 2 dx Step 3: Simplify the left-hand side and integrate: Z Z 1 dy = 1 dx y2 R Step 4: Integrate y12 dy on the left-hand side: − 1 =x+C y where C is the constant of integration. Step 5: Solve for y by multiplying both sides by -1: 1 = −x − C y 6 Step 6: Apply the initial condition y(0) = 1 to find the value of the constant C: 1 = −0 − C =⇒ C = −1 1 Step 7: Substitute the value of C back into the equation: 1 = −x + 1 y Step 8: Solve for y: y= 1 −x + 1 Therefore, the particular solution to the initial value problem is y = 1 1−x . Question 7 Question Consider the initial value problem: x+y dy = , dx x−y y(0) = 1. Find the particular solution to the initial value problem on the interval (−∞, ∞). Solution To solve equation solve it. Step equation this initial value problem, we will first write the given differential in a more standard form and then proceed to separate variables to 1: Rewrite the differential equation. We rewrite the given differential as: 1 + xy dy x+y dy = =⇒ = . dx x−y dx 1 − xy Step 2: Separate variables. Separating variables in the differential equation, we get: 1 − xy x−y dy = dx =⇒ dy = dx. 1 + xy x+y Step 3: Integrate both sides. Integrating both sides of the equation, we have: Z Z x−y dy = dx. x+y Step 4: Perform the integration. Integrating the left side, we get: Z Z Z x−y x y dy = dy − dy. x+y x+y x+y 7 Solving these integrals and simplifying, we obtain: x ln |x + y| − y ln |x + y| = x + C, where C is the constant of integration. Step 5: Apply the initial condition. Using the initial condition y(0) = 1, we substitute x = 0 and y = 1 into the equation above: 0 ln |0 + 1| − 1 ln |0 + 1| = 0 + C. This simplifies to: − ln(1) = C =⇒ C = 0. Step 6: Final solution. Substitute C = 0 back into the solution to get the particular solution: x ln |x + y| − y ln |x + y| = x. Question 8 Question Solve the initial value problem: y ′ = 2y + 3x with y(0) = 1. Solution Step 1: First, we need to find the general solution to the differential equation y ′ = 2y + 3x. This is a first-order linear differential equation, whichR can be dy −2y = 3x. The integrating factor is e −2dx = written in the standard form as dx −2x e . Step 2: Multiply both sides of the standard form by the integrating factor dy e−2x to get: e−2x dx − 2e−2x y = 3xe−2x . d Step 3: Rewrite the left-hand side using the product rule: dx (e−2x y) = −2x 3xe . R d −2x R Step 4: Integrate both sides with respect to x: dx (e y) dx = 3xe−2x dx. This gives e−2x y = − 23 xe−2x + C, where C is the constant of integration. Step 5: Solve for y by multiplying by e2x : y = − 23 x + Ce2x . Step 6: Apply the initial condition y(0) = 1 to find the value of C: 1 = − 23 (0) + Ce2(0) . So, C = 1. Step 7: Substitute C = 1 back into the general solution to obtain the particular solution: y = − 32 x + e2x . Therefore, the solution to the initial value problem y ′ = 2y+3x with y(0) = 1 is y = − 23 x + e2x . 8 Question 9 Question Solve the initial value problem: y ′ + y = e−2x , y(0) = 3 Solution Step 1: First, let’s find the integrating factor µ(x) by multiplying both sides of the differential equation by it: µ(x) = e R 1 dx = ex Step 2: Now, multiply the differential equation by the integrating factor: ex y ′ + ex y = e−x · e−2x Step 3: Notice that the left side can be rewritten as the derivative of the product ex y: (ex y)′ = e−3x Step 4: Integrate both sides with respect to x: Z Z x ′ (e y) dx = e−3x dx Step 5: Solve the integrals: 1 ex y = − e−3x + C 3 Step 6: Next, solve for y by dividing by ex : 1 y = − e−4x + Ce−x 3 Step 7: Use the initial condition y(0) = 3 to find the value of C: 1 3=− +C 3 ⇒ C= 10 3 Step 8: Finally, the solution to the initial value problem is: 1 10 y = − e−4x + e−x 3 3 9 Question 10 Question Consider the initial value problem dy = x2 + y, dx y(0) = 1. Find the particular solution to the initial value problem using the method of integrating factors. Solution Step 1: First, we rewrite the given differential equation in the form y ′ +P (x)y = Q(x) where P (x) and Q(x) are functions of x. dy = x2 + y dx dy − y = x2 . dx Step 2: Next, we identify P (x) and Q(x): Q(x) = x2 . P (x) = −1, e R Step 3: To find the integrating factor I(x), we use the formula I(x) = . P (x) dx I(x) = e R −1 dx = e−x . Step 4: Multiply the integrating factor by the differential equation. e−x dy − e−x y = x2 e−x dx d −x (e y) = x2 e−x . dx Step 5: Integrate both sides with respect to x. Z Z d −x (e y) dx = x2 e−x dx dx e−x y = (−x2 − 2x − 2)e−x + C, where C is the constant of integration. Step 6: Solve for y. y = −x2 − 2x − 2 + Cex . 10 Step 7: To find the particular solution, we substitute x = 0 and y = 1 into the equation. 1 = −02 − 2(0) − 2 + Ce0 1 = −2 + C C = 3. Therefore, the particular solution to the initial value problem is y = −x2 − 2x − 2 + 3ex . Question 11 Question Consider the initial value problem: x2 + y 2 dy = , dx xy y(1) = 2 Use the method of separation of variables to find the particular solution to this initial value problem. Solution Step 1: Rewrite the differential equation by separating variables: x2 + y 2 y x dy = =⇒ 2 dy = 2 dx dx xy y x Step 2: Integrate both sides with respect to their respective variables: Z Z 1 1 dy = dx y x Step 3: Solve the integrals: ln |y| = ln |x| + C where C is the constant of integration. Step 4: Exponentiate both sides to remove the natural logarithm: y = eln |x|+C = eln |x| · eC = Celn |x| = C|x| Step 5: Apply the initial condition y(1) = 2 to find the value of the constant C: 2 = C · 1 =⇒ C = 2 Step 6: Substitute C = 2 back into the general solution to find the particular solution: y = 2|x| Therefore, the particular solution to the initial value problem is y = 2|x|. 11 Question 12 Question Solve the initial value problem: dy + 2y = 4x, dx y(0) = 3 Solution R Step 1: First, we identify the integrating factor, which is given by e 2 dx = e2x . Step 2: Multiply the integrating factor by the differential equation: e2x dy + 2e2x y = 4xe2x dx Step 3: Recognize that the left-hand side is the derivative of the product of y and the integrating factor: d (ye2x ) = 4xe2x dx Step 4: Integrate with respect to x: Z Z d 2x (ye ) dx = 4xe2x dx dx Step 5: Solve the integrals: ye2x = Z 4xe2x dx ye2x = 2xe2x − Z 2e2x dx ye2x = 2xe2x − e2x + C1 Step 6: Solve for y: y = 2x − 1 + C1 e−2x Step 7: Apply the initial condition y(0) = 3 to determine the value of C1 : 3 = 2(0) − 1 + C1 C1 = 4 Step 8: The solution to the initial value problem is: y = 2x − 1 + 4e−2x 12 Question 13 Question Consider the initial value problem dy = 2x(3 − y 2 ), dx y(0) = 1. Find the particular solution to this initial value problem. Solution Step 1: First, let’s separate variables in the given differential equation. dy = 2x dx. 3 − y2 Step 2: Next, we integrate both sides of the equation. Z Z 1 dy = 2 x dx. 3 − y2 Step 3: To integrate the left side, we use partial fraction decomposition. Let A and B be constants such that 1 A B =√ +√ . 2 3−y 3+y 3−y Step 4: Multiplying both sides by 3 − y 2 gives √ √ 1 = A( 3 − y) + B( 3 + y). √ √ Step 5: Set y = 3 to find A and set y = − 3 to find B. Solving the system 1 1 of equations gives A = − 2√ and B = 2√ . 3 3 Step 6: Now, we can rewrite the integral as Z Z Z 1 1 1 1 √ √ − √ dy + √ dy = 2 x dx. 2 3 3+y 2 3 3−y Step 7: After integrating both sides, we get √ √ 1 1 − √ ln | 3 + y| + √ ln | 3 − y| = x2 + C, 2 3 2 3 where C is the constant of integration. Step 8: Simplifying the left side and using the initial condition y(0) = 1, we find C = 0. Step 9: Thus, the particular solution to the initial value problem is √ √ 1 1 − √ ln | 3 + y| + √ ln | 3 − y| = x2 . 2 3 2 3 13 Question 14 Question Solve the initial value problem: dy 2y − x = , dx x+y y(0) = 1. Solution Step 1: Rewrite the differential equation in a more standard form by dividing both sides by x + y: 2y − x dy 2y/x − 1 dy = =⇒ = . dx x+y dx 1 + y/x Step 2: Let u = y/x, then y = ux and the differential equation gives: u+x dy dx = u + x du dx . Substituting this into du 2u − 1 du u−1 = =⇒ x = . dx 1+u dx 1+u Step 3: Separate variables and integrate both sides: Z Z 1+u 1 du = dx. u−1 x Step 4: Solve the integral on the left side: Z Z Z 1+u (u − 1) + 2 2 du = du = 1 + du = u + 2 ln |u − 1| + C. u−1 u−1 u−1 Step 5: Substitute back u = y/x and solve for y: y + 2 ln |y/x − 1| = ln |x| + C. x Step 6: Use the initial condition y(0) = 1 to find the value of C: 1 + 2 ln |1 − 1| = ln |0| + C =⇒ 0 = −∞ + C =⇒ C = ∞. 0 Step 7: The particular solution is then: y + 2 ln |y/x − 1| = ln |x| + ∞. x Therefore, the solution to the initial value problem is ln |x| + ∞. 14 y x + 2 ln |y/x − 1| = Question 15 Question Consider the initial value problem: dy 2x = , dx y y(0) = 1 Solution Step 1: Rewrite the initial value problem as a separable first-order differential equation. y dy = 2x dx Step 2: Integrate both sides of the equation. Z Z y dy = 2x dx Step 3: Solve the integrals. 1 2 y = x2 + C 2 where C is the constant of integration. Step 4: Use the initial condition y(0) = 1 to find the value of the constant C. 1 1 2 · 1 = 02 + C =⇒ C = 2 2 Step 5: Substitute the value of C back into the equation. 1 2 1 y = x2 + 2 2 Step 6: Solve for y in terms of x. y 2 = 2x2 + 1 ⇒ p y = ± 2x2 + 1 √ Therefore, the solution to the initial value problem is y = ± 2x2 + 1, but √ since y(0) = 1, the solution is y = 2x2 + 1. Question 16 Question Solve the initial value problem: 1 dy = sin(2x) + y, dx 2 15 y(0) = 1 Solution Step 1: Rewrite the initial value problem: dy 1 = sin(2x) + y dx 2 with y(0) = 1 Step 2: This is a first-order linear differential equation in the form p(x)y = q(x), where p(x) = 12 and q(x) = sin(2x). R Step 3: We start by finding the integrating factor I(x) = e p(x)dx : R I(x) = e 1 2 dx dy dx + 1 = e2x Step 4: Multiply both sides of the differential equation by the integrating factor I(x): 1 1 1 dy 1 e2x + e 2 x · y = e 2 x sin(2x) dx 2 Step 5: Rewrite the left-hand side as the derivative of the product I(x)y using the product rule: 1 d  1x  e 2 y = e 2 x sin(2x) dx Step 6: Integrate both sides with respect to x: Z Z 1 d  1x  e 2 y dx = e 2 x sin(2x)dx dx Step 7: Integrating the right-hand side with respect to x, we get: 1 1 1 e 2 x y = − e 2 x cos(2x) + C 2 1 Step 8: Solve for y by multiplying through by e− 2 x : 1 1 y = − cos(2x) + Ce− 2 x 2 Step 9: Use the initial condition y(0) = 1 to solve for the constant C: 1 1 1 3 1 = − cos(0) + Ce− 2 ·0 =⇒ 1 = − + C =⇒ C = 2 2 2 Step 10: Substitute the value of C back into the general solution to find the particular solution: 1 3 1 y = − cos(2x) + e− 2 x 2 2 Therefore, the solution to the initial value problem 1 y(0) = 1 is y = − 12 cos(2x) + 32 e− 2 x . 16 dy dx = sin(2x) + 12 y with Question 17 Question Solve the initial value problem: y ′ = 4y − e5x , y(0) = 3 Solution Step 1: First, we find the general solution to the given differential equation. The standard form of the linear first-order differential equation is y ′ + P (x)y = Q(x). Comparing the given equation y ′ = 4y − e5x with the standard form, we have P (x) = 4 and Q(x) = e5x . The integrating factor I(x) is given by: I(x) = e R P (x) dx =e R 4 dx = e4x Multiplying both sides of the differential equation by the integrating factor, we get: e4x y ′ + 4e4x y = e9x Now, we can rewrite the left side of the above equation as the derivative of the product of e4x and y using the product rule: d 4x (e y) = e9x dx Integrating both sides with respect to x gives us: e4x y = 1 9x e +C 9 (Where C is the constant of integration) Step 2: Next, we apply the initial condition y(0) = 3 to find the value of the constant C: 1 e4(0) · 3 = e9(0) + C 9 1 3= +C 9 1 26 C =3− = 9 9 Therefore, the particular solution to the initial value problem is: e4x y = y= 1 9x 26 e + 9 9 1 5x 26 −4x e + e 9 9 17 Question 18 Question Consider the initial value problem given by dy x+y = , dx x−y y(0) = 1. Find the particular solution to the initial value problem. Solution Step 1: Rewrite the initial value problem in a separable form. x+y dy = dx x−y dy x+y = dx x−y (x − y) dy = (x + y) dx Step 2: Integrate both sides to solve for y. Z Z (x − y) dy = (x + y) dx Z Z Z Z x dy − y dy = x dx + y dx x2 y2 y2 = + + C1 2 2 2 where C1 is the constant of integration. Step 3: Use the initial condition y(0) = 1 to find the value of C1 . xy − (0)(1) − (1)2 (0)2 (1)2 = + + C1 2 2 2 1 1 − = + C1 2 2 −1 = 1 + 2C1 C1 = −1 Step 4: Substitute C1 = −1 back into the equation to find the particular solution. y2 x2 y2 xy − = + −1 2 2 2 2xy − 2y 2 = x2 + y 2 − 2 2xy − 2y 2 = x2 + y 2 − 2 2xy + y 2 = x2 + 2 Therefore, the particular solution to the initial value problem is 2xy + y 2 = x + 2. 2 18 Question 19 Question Consider the initial value problem: dy = 2e3x − 5y, dx y(0) = 3 Solution Step 1: First we rewrite the differential equation as a separable equation: dy + 5y = 2e3x dx Step 2: To solve the separable equation, we separate variables: dy dy + 5y = 2e3x =⇒ = 2e3x − 5y dx dx Step 3: We now solve the separable equation: Separating variables, we have: dy = dx 2e3x − 5y Step 4: Next, we integrate both sides of the equation: Z Z 1 dy = dx 2e3x − 5y Step 5: By substituting u = 2e3x − 5y, du = −5dy, we get: Z Z −1 du = dx 5u Step 6: Integrating both sides, we have: −1 ln |u| = x + C 5 Step 7: Substituting back u = 2e3x − 5y and rearranging, we get: ln |2e3x − 5y| = −5x − 5C Step 8: Next, we solve for y: 2e3x − 5y = e−5x−5C Step 9: Applying the initial condition y(0) = 3, we can solve for C: 2e0 − 5(3) = e0 =⇒ 2 − 15 = 1 =⇒ C = −13 Step 10: Substituting this value of C back in, we have: 2e3x − 5y = e−5x−5(−13) Step 11: Simplifying, we get the solution to the initial value problem: 2e3x − 5y = e−5x+65 19 Question 20 Question Solve the initial value problem: dy x = , dx y y(0) = 1 Solution dy = xy . Step 2: Rearrange the Step 1: Rewrite the differential equation as dx terms to separate variables: y dy = x dx. Step 3: Integrate both sides of the equation: Z Z y dy = x dx Step 4: Solve the integrals to get: y2 x2 = +C 2 2 where C is the constant of integration. Step 5: Find the value of the constant C using the initial condition y(0) = 1: 12 02 1 = + C =⇒ =C 2 2 2 So, the equation becomes y2 2 = x2 2 + 21 . Step 6: Solve for y in terms of x: p y 2 = x2 + 1 =⇒ y = ± x2 + 1 √ Step 7: Since y(0) = 1, we choose y = x2 + 1. Therefore, the solution to the √ initial value problem is y = x2 + 1. Question 21 Question Consider the initial value problem: dy = 2y + x, dx y(0) = 1 Find the particular solution to the differential equation that satisfies the initial condition. 20 Solution dy Step 1: First, let’s solve the differential equation dx = 2y +R x using the method of integrating factors. The integrating factor is given by e 2dx = e2x . d 2x (e y) = e2x (2y + x) dx Z e2x y = e2x (2y + x)dx Z Z e2x y = 2e2x ydx + xe2x dx Step 2: We can solve the integrals on the right side of the equation to get: Z Z 2x 2x e y = 2 e ydx + xe2x dx Z Z 1 1 e2x dx e2x y = 2e2x y − e2x dy + xe2x − 2 2 1 1 e2x y = 2e2x y − e2x + xe2x − e2x + C 2 4 Step 3: Now, applying the initial condition y(0) = 1, we get: e0 · 1 = 2e0 · 1 − e0 + 1=2−1− 1 1 · 0 · e0 − · e0 + C 2 4 1 4 1 +C 4 3 C= 4 1= Step 4: Substituting C = 3 4 back into our equation, we find: 1 1 3 e2x y = 2e2x y − e2x + xe2x − e2x + 2 4 4 1 2x 1 2x 3 2x 2x 2x e y − 2e y = −e + xe − e + 2 4 4 1 2x 1 2x 3 2x 2x −e y = −e + xe − e + 2 4 4 1 1 3 −2x y =1− x+ + e 2 4 4 Therefore, the particular solution to the differential equation that satisfies the initial condition y(0) = 1 is y = 1 − 12 x + 14 + 34 e−2x . 21 Question 22 Question Consider the initial value problem: dy = 2x + 3y, dx y(0) = 1. Find the particular solution to the initial value problem. Solution Step 1: Rewrite the given initial value problem as: dy − 3y = 2x, dx y(0) = 1. dy Step 2: This is a first-order linear differential Requation of the form dx + P (x)dx P (x)y = Q(x). The integrating factor is given by e . R Step 3: In this case, P (x) = −3, so the integrating factor is e −3dx = e−3x . Step 4: Multiply both sides of the differential equation by the integrating factor: dy − 3e−3x y = 2xe−3x . e−3x dx Step 5: Rewrite the equation as the derivative of a product: d (ye−3x ) = 2xe−3x . dx Step 6: Integrate both sides with respect to x: Z Z d (ye−3x ) dx = 2xe−3x dx. dx Step 7: This simplifies to: 2 2 ye−3x = − xe−3x − e−3x + C, 3 9 where C is the constant of integration. Step 8: Applying the initial condition y(0) = 1, we have: 1=0− 2 +C 9 ⇒ C= 11 . 9 Step 9: Substitute C = 11 9 back into the equation to find the particular solution: 2 11 2 ye−3x = − xe−3x − e−3x + . 3 9 9 Thus, the particular solution to the initial value problem is y = − 23 x − 92 + 11 3x 9 e . 22 Question 23 Question Solve the initial value problem: y ′ = 3y − 6, y(0) = 2 Solution Step 1: First, let’s solve the differential equation y ′ = 3y − 6 by separating variables: dy = 3y − 6 dt dy = dt 3y − 6 Step 2: Integrate both sides with respect to y and t: Z Z 1 dy = dt 3y − 6 1 ln |3y − 6| = t + C1 3 ln |3y − 6| = 3t + C2 |3y − 6| = e3t+C2 |3y − 6| = Ce3t 3y − 6 = Ce3t (where C = eC2 ) 3y = Ce3t + 6
1 Ce3t + 6 y= 3 Step 3: Next, we apply the initial condition y(0) = 2 to find the value of C: 2= or
1 C · e0 + 6 3 1 (C + 6) 3 6=C +6 2= C=0 Step 4: Now that we have C = 0, the solution to the initial value problem is: y= 1 · 6 = 2e3t 3 23 Question 24 Question Solve the initial value problem: dy 3x2 = , dx y y(1) = 2 Solution Step 1: Rewrite the differential equation in a separable form. y dy = 3x2 dx Step 2: Integrate both sides with respect to their respective variables. Z Z y dy = 3x2 dx Step 3: Solve the integrals. 1 2 y = x3 + C 2 where C is the constant of integration. Step 4: Apply the initial condition y(1) = 2 to find the value of C. 1 2 (2) = 13 + C =⇒ 2 = 1 + C =⇒ C = 1 2 Step 5: Substitute the value of C back into the equation. 1 2 y = x3 + 1 2 Step 6: Solve for y in terms of x. y 2 = 2x3 + 2 =⇒ y = ± p 2x3 + 2 Therefore, the solution to the initial value problem is y = y(1) = 2. Question 25 Question Solve the initial value problem: dy = 2x + y, dx 24 y(0) = 1 √ 2x3 + 2 since Solution Step 1: Write the given initial value problem as: dy − y = 2x, y(0) = 1 dx Step 2: This is a first-order linear differential equation.R To solve it, we first find the integrating factor µ(x) which is given by µ(x) = e −1 dx . µ(x) = e R −1 dx = e−x Step 3: Multiply both sides of the differential equation by the integrating factor µ(x): dy − e−x y = 2xe−x e−x dx Step 4: Recognize the left side as the derivative of the product of e−x and y (this step involves integrating factors): d (ye−x ) = 2xe−x dx Step 5: Integrate both sides with respect to x: Z Z d −x (ye ) dx = 2xe−x dx dx Step 6: Integrate to get: ye−x = −2xe−x − 2e−x + C Step 7: Solve for y by multiplying both sides by ex : y = −2x − 2 + Cex Step 8: Use the initial condition y(0) = 1 to find C: 1 = −2(0) − 2 + C C=3 Step 9: Substitute C = 3 back into the general solution to get the particular solution: y = −2x − 2 + 3ex Therefore, the solution to the initial value problem is y = −2x − 2 + 3ex with the initial condition y(0) = 1. Question 26 Question Solve the initial value problem: y ′ = 2xy 2 , 25 y(0) = 1. Solution Step 1: Let’s rewrite the differential equation in a form suitable for separation of variables: dy = 2xy 2 . dx Step 2: Separate variables by dividing both sides by y 2 and multiplying by dx: dy = 2xdx. y2 Step 3: Integrate both sides: Z 1 dy = 2 y2 Z xdx. Step 4: Solving the integrals gives: − 1 = x2 + C, y where C is the constant of integration. Step 5: To find the value of the constant C, we use the initial condition y(0) = 1: 1 − = 02 + C ⇒ C = −1. 1 Step 6: Substituting C back into the equation gives: − 1 = x2 − 1. y Step 7: Solve for y: y= 1 . 1 − x2 Therefore, the solution to the initial value problem is y = initial condition y(0) = 1. Question 27 Question Solve the initial value problem: dy 2x = 2, dx y 26 y(0) = 1 1 1−x2 with the Solution Step 1: Separate variables by writing the given differential equation as: y 2 dy = 2xdx Step 2: Integrate both sides with respect to their respective variables: Z Z y 2 dy = 2xdx Step 3: Integrate to solve for y: y3 = x2 + C 3 where C is the constant of integration. Step 4: Applying the initial condition y(0) = 1 to solve for C: 1 13 = 0 + C =⇒ C = 3 3 Step 5: Substitute the value of C back into the equation to find the particular solution: y3 1 = x2 + =⇒ y 3 = 3x2 + 1 3 3 Step 6: Solve for y by taking the cube root of both sides: p 3 y = 3x2 + 1 √ Therefore, the solution to the initial value problem is y = 3 3x2 + 1 with the initial condition y(0) = 1. Question 28 Question Consider the initial value problem: 2x dy = , dx y y(0) = 2. Solution Step 1: Rewrite the differential equation in a separable form by isolating variables x and y: y dy = 2x dx. Step 2: Integrate both sides of the equation: Z Z y dy = 2x dx. 27 Step 3: Solve the integrals: y2 = x2 + C, 2 where C is the constant of integration. Step 4: Use the initial condition y(0) = 2 to find the value of the constant C: 22 = 0 + C =⇒ C = 2. 2 Step 5: Substitute the value of C back into the equation: y2 = x2 + 2. 2 Step 6: Solve for y in terms of x: p y 2 = 2(x2 + 2) =⇒ y = ± 2x2 + 4. Therefore, the solution to the initial value problem is y = y(0) = 2. √ 2x2 + 4 since Question 29 Question Solve the initial value problem: dy 4x + 3y − 6 = , dx 3x + 2y − 5 y(0) = 2 Solution Step 1: First, we rewrite the given differential equation in terms of dy and dx by multiplying both sides by dx: dx · dy 4x + 3y − 6 = · dx dx 3x + 2y − 5 Step 2: To make it easier to solve, we rewrite the equation as follows: 3x + 2y − 5 dy = dx 4x + 3y − 6 Step 3: We now integrate both sides, and find a constant of integration, C: Z Z 3x + 2y − 5 dy = dx 4x + 3y − 6 28 Step 4: After integrating, we get: 1 ln |4x + 3y − 6| = x + C 2 Step 5: We can rewrite the expression using exponential functions: |4x + 3y − 6| = e2x+2C Step 6: Since y(0) = 2, we substitute x = 0 and y = 2 into the equation: |4(0) + 3(2) − 6| = e2(0)+2C Step 7: Simplifying, we get: |6| = e2C Step 8: Thus, 6 = e2C or e2C = 6. Taking the natural logarithm of both sides, we get: 2C = ln 6 Step 9: Solving for C, we have: C= 1 ln 6 2 Step 10: Substitute the value of C back into the equation we obtained: |4x + 3y − 6| = e2x+ln 6 Step 11: Finally, we simplify the equation to get the solution to the initial value problem: 4x + 3y − 6 = 6e2x Question 30 Question Consider the initial value problem: dy = 2x3 + 3x2 + x − 1, dx y(0) = 5 Use Euler’s method with step size h = 0.1 to approximate the value of y(0.2). 29 Solution Step 1: Define the step size h, the initial condition, and the function f (x, y). ˆ Step size: h = 0.1 ˆ Initial condition: y(0) = 5 ˆ Function f (x, y) = 2x3 + 3x2 + x − 1 Step 2: Perform Euler’s method to approximate y(0.1), y(0.2), . . . until x = 0.2. x1 = x0 + h = 0 + 0.1 = 0.1 y1 = y0 + hf (x0 , y0 ) = 5 + 0.1(2(0)3 + 3(0)2 + 0 − 1) = 5 − 0.1 = 4.9 x2 = x1 + h = 0.1 + 0.1 = 0.2 y2 = y1 + hf (x1 , y1 ) = 4.9 + 0.1(2(0.1)3 + 3(0.1)2 + 0.1 − 1) = 4.9 + 0.1(0.002 + 0.03 + 0.1 − 1) = 4.9 + 0.1(−0.968) = 4.9 − 0.0968 = 4.8032 Therefore, the approximate value of y(0.2) using Euler’s method with step size h = 0.1 is y(0.2) ≈ 4.8032. Question 31 Question Solve the initial value problem: 2x dy = , dx y y(0) = 1 Solution Step 1: Rewrite the differential equation in terms of y and dy: y dy = 2x dx Step 2: Integrate both sides: Z Z y dy = 2x dx y2 = x2 + C1 2 30 Step 3: Apply the initial condition y(0) = 1: 12 = 0 + C1 2 1 C1 = 2 Step 4: Substitute C1 = 1 2 back into the equation: y2 1 = x2 + 2 2 y 2 = 2x2 + 1 Step 5: Solve for y by taking the square root: p y = ± 2x2 + 1 Step 6: Apply the initial condition y(0) = 1 to determine the specific solution: p 1 = 2(0)2 + 1 √ 1= 1 y=1 Therefore, the solution to the initial value problem is y = 1. Question 32 Question Solve the initial value problem: dy dx = y 2 + x2 , y(0) = 1. Solution Step 1: Rewrite the equation as a separable first-order differential equation by bringing all terms involving y to the left side and all terms involving x to the right side. dy = y 2 + x2 dx dy = dx y2 Step 2: Integrate both sides with respect to y and x separately. Z Z 1 dy = dx y2 − 1 = x + C1 y 31 where C1 is the constant of integration. Step 3: Solve for y in terms of x by isolating y. − 1 = x + C1 y −1 = y(x + C1 ) y= 1 1 − x − C1 Step 4: Use the initial condition y(0) = 1 to find the value of the constant C1 . 1 1 − 0 − C1 1 1= 1 − C1 1= 1 − C1 = 1 C1 = 0 Step 5: Substitute C1 = 0 back into the expression for y to obtain the particular solution. 1 y= 1−x Therefore, the solution to the initial value problem is y = 1 1−x . Question 33 Question Solve the initial value problem: dy = 2x(1 − y), dx y(0) = 0 Solution Step 1: First, rewrite the differential equation in a separable form by moving all terms involving y to one side and all terms involving x to the other side: dy = 2x dx 1−y Step 2: Integrate both sides with respect to their respective variables: Z Z 1 dy = 2x dx 1−y 32 Step 3: Solve the integrals: − ln |1 − y| = x2 + C1 where C1 is the constant of integration. Step 4: Exponentiate both sides to eliminate the natural logarithm: |1 − y| = e−x 2 −C1 = Ce−x 2 where C = ±e−C1 . Step 5: Rewrite the equation without absolute values: 1 − y = Ce−x 2 y = 1 − Ce−x 2 or Step 6: Apply the initial condition y(0) = 0 to solve for the constant C: 0 = 1 − Ce−0 C=1 Step 7: Substitute C = 1 back into the solution: y = 1 − e−x 2 2 Therefore, the solution to the initial value problem is y = 1 − e−x . Question 34 Question Solve the initial value problem: 1 dy = , dx x+y y(1) = 1 Solution Step 1: Rewrite the differential equation in standard form by separating variables: (1 + y) dy = dx Step 2: Integrate both sides with respect to their respective variables: Z Z (1 + y) dy = dx Z Z 1 dy + Z y dy = 33 dx y2 =x+C 2 Step 3: Use the initial condition y(1) = 1 to find the value of the constant y+ C: 1+ Step 4: Substitute C = solution: 1 2 1 =1+C 2 1 C= 2 back into the equation to find the particular y2 1 =x+ 2 2 Step 5: Use the quadratic formula to solve for y: y+ y 2 + 2y − 2x − 1 = 0 p −2 ± 22 − 4 · 1 · (−2x − 1) y= 2·1 √ −2 ± 4 + 8x + 4 y= 2 √ −2 ± 8x + 8 y= 2 √ y = −1 ± 2x + 2 Step 6: Use the initial condition y(1) = 1 to determine the specific solution: √ 1 = −1 ± 2 · 1 + 2 √ 2= 4 √ Since 4 = 2 > 1, we choose the positive sign: √ y = −1 + 2x + 2 √ Therefore, the solution to the initial value problem is y = −1 + 2x + 2. Question 35 Question Consider the initial value problem: dy = 2y, dx y(0) = 3 Find the solution to this initial value problem. 34 Step 4: Solve the integral on the right-hand side: 2 2 ye−3x = e−3x (− x − ) + C 3 9 2 2 y = − x − + Ce3x 3 9 Step 5: Use the initial condition y(0) = 1 to find the value of C: 2 1=− +C 9 C =1+ Step 6: Substitute C = 11 9 11 2 = 9 9 back into the particular solution: 2 2 11 y = − x − + e3x 3 9 9 2 9 Therefore, the particular solution to the initial value problem is y = − 32 x − 3x + 11 9 e . Question 2 Question Solve the initial value problem: y ′ + 2y = 4e−2x , y(0) = 3 Solution Step 1: First, we find the integrating factor: Let P (x) = 2, then the integrating factor is given by: R R µ(x) = e P (x) dx = e 2 dx = e2x Step 2: Next, we multiply the given equation by the integrating factor µ(x): e2x y ′ + 2e2x y = 4e2x e−2x e2x y ′ + 2e2x y = 4 Step 3: Rewrite the left side of the equation as the derivative of the product y(x)e2x : (e2x y)′ = 4 Step 4: Integrate both sides with respect to x: Z Z 2x ′ (e y) dx = 4 dx 2 e2x y = 4x + C Step 5: Apply the initial condition y(0) = 3 to find the value of the constant C: e2·0 · 3 = 4 · 0 + C 3=C Step 6: Write the final solution by substituting the value of C back into the equation: e2x y = 4x + 3 y= 4x + 3 4x + 3 = 2x e e2x Question 3 Question Solve the initial value problem: y ′ = 2y − 4 with the initial condition y(0) = 3. Solution Step 1: First, we solve the differential equation y ′ = 2y − 4. This is a firstorder linear differential equation. To solve it, we rewrite it in the standard form y ′ − 2y = −4. Step 2: To solve the homogeneous part of the R equation, we find the integrating factor. The integrating factor is given by e −2 dx = e−2x . Step 3: Multiply the integrating factor by the given equation to get e−2x y ′ − 2e−2x y = −4e−2x . d Step 4: Rewriting the left side as the derivative of e−2x y, we have dx (e−2x y) = −2x −4e . Step 5: Integrating both sides with respect to x gives e−2x y = 2e−2x + C, where C is the constant of integration. Step 6: Solving for y, we have y = 2 + Ce2x . Step 7: To find the particular solution, we use the initial condition y(0) = 3. Substitute x = 0 and y = 3 into the equation y = 2 + Ce2x to find the value of C. Step 8: We have 3 = 2 + C, which implies C = 1. Step 9: Substituting C = 1 back into the equation y = 2 + Ce2x gives the particular solution y = 2 + e2x . Therefore, the solution to the initial value problem y ′ = 2y − 4 with y(0) = 3 is y = 2 + e2x . 3 Question 4 Question Solve the initial value problem: y ′ + 2xy = 2x, y(0) = 1 Solution Step 1: First, we find the integrating factor by calculating e Z 2xdx = x2 R 2xdx . 2 Therefore, the integrating factor is ex . Step 2: Multiply both sides of the differential equation by the integrating factor and rewrite the left-hand side as a product rule: 2 2 ex y ′ + 2xex y = 2xex 2 Step 3: Recognize that the left-hand side is now the derivative of a product: 2 (ex y)′ = 2xex 2 Step 4: Integrate both sides with respect to x: Z Z 2 x2 ′ (e y) dx = 2xex dx Z x2 e y= 2 2xex dx + C Step 5: Solve the integral on the right-hand side: 2 2 ex y = ex + C 2 Step 6: Solve for y by dividing by ex : y = 1 + Ce−x 2 Step 7: Use the initial condition y(0) = 1 to find the value of the constant C: 1 = 1 + C · e0 C=0 Step 8: Substitute the value of C back into the equation to obtain the particular solution: y=1 Therefore, the solution to the initial value problem is y = 1. 4 Question 5 Question Solve the initial value problem: dy = 3y − 6x, dx y(0) = 4 Solution Step 1: First, we identify the differential equation as a first-order linear equation and rewrite it in standard form dy/dx + P (x)y = Q(x), where P (x) and Q(x) are functions of x. dy = 3y − 6x dx dy − 3y = −6x dx Step 2: Next, we identify P (x) = −3 and Q(x) = −6x. We now find the R integrating factor I(x) = e P (x)dx . I(x) = e R −3dx = e−3x Step 3: We multiply the differential equation by the integrating factor I(x). e−3x dy − 3e−3x y = −6xe−3x dx d −3x (e y) = −6xe−3x dx Step 4: Integrate both sides of the equation with respect to x. Z Z d −3x (e y)dx = −6xe−3x dx dx Z −3x e y = −6xe−3x dx Step 5: Solve the integral on the right side. Z −3x −3x e y = 3xe − 3e−3x dx e−3x y = 3xe−3x + C Step 6: Multiply through by e3x . y = 3x + Ce3x 5 Step 7: Apply the initial condition y(0) = 4 to solve for C. 4 = 3(0) + C(1) C=4 Step 8: Substitute C = 4 back into the general solution. y = 3x + 4e3x Therefore, the solution to the initial value problem is y = 3x + 4e3x . Question 6 Question Consider the initial value problem: dy = y2 , dx y(0) = 1. Find the particular solution to the differential equation that satisfies the initial condition. Solution Step 1: Separate the variables: 1 dy =1 y 2 dx Step 2: Integrate both sides with respect to x: Z Z 1 dy dx = 1 dx y 2 dx Step 3: Simplify the left-hand side and integrate: Z Z 1 dy = 1 dx y2 R Step 4: Integrate y12 dy on the left-hand side: − 1 =x+C y where C is the constant of integration. Step 5: Solve for y by multiplying both sides by -1: 1 = −x − C y 6 Step 6: Apply the initial condition y(0) = 1 to find the value of the constant C: 1 = −0 − C =⇒ C = −1 1 Step 7: Substitute the value of C back into the equation: 1 = −x + 1 y Step 8: Solve for y: y= 1 −x + 1 Therefore, the particular solution to the initial value problem is y = 1 1−x . Question 7 Question Consider the initial value problem: x+y dy = , dx x−y y(0) = 1. Find the particular solution to the initial value problem on the interval (−∞, ∞). Solution To solve equation solve it. Step equation this initial value problem, we will first write the given differential in a more standard form and then proceed to separate variables to 1: Rewrite the differential equation. We rewrite the given differential as: 1 + xy dy x+y dy = =⇒ = . dx x−y dx 1 − xy Step 2: Separate variables. Separating variables in the differential equation, we get: 1 − xy x−y dy = dx =⇒ dy = dx. 1 + xy x+y Step 3: Integrate both sides. Integrating both sides of the equation, we have: Z Z x−y dy = dx. x+y Step 4: Perform the integration. Integrating the left side, we get: Z Z Z x−y x y dy = dy − dy. x+y x+y x+y 7 Solving these integrals and simplifying, we obtain: x ln |x + y| − y ln |x + y| = x + C, where C is the constant of integration. Step 5: Apply the initial condition. Using the initial condition y(0) = 1, we substitute x = 0 and y = 1 into the equation above: 0 ln |0 + 1| − 1 ln |0 + 1| = 0 + C. This simplifies to: − ln(1) = C =⇒ C = 0. Step 6: Final solution. Substitute C = 0 back into the solution to get the particular solution: x ln |x + y| − y ln |x + y| = x. Question 8 Question Solve the initial value problem: y ′ = 2y + 3x with y(0) = 1. Solution Step 1: First, we need to find the general solution to the differential equation y ′ = 2y + 3x. This is a first-order linear differential equation, whichR can be dy −2y = 3x. The integrating factor is e −2dx = written in the standard form as dx −2x e . Step 2: Multiply both sides of the standard form by the integrating factor dy e−2x to get: e−2x dx − 2e−2x y = 3xe−2x . d Step 3: Rewrite the left-hand side using the product rule: dx (e−2x y) = −2x 3xe . R d −2x R Step 4: Integrate both sides with respect to x: dx (e y) dx = 3xe−2x dx. This gives e−2x y = − 23 xe−2x + C, where C is the constant of integration. Step 5: Solve for y by multiplying by e2x : y = − 23 x + Ce2x . Step 6: Apply the initial condition y(0) = 1 to find the value of C: 1 = − 23 (0) + Ce2(0) . So, C = 1. Step 7: Substitute C = 1 back into the general solution to obtain the particular solution: y = − 32 x + e2x . Therefore, the solution to the initial value problem y ′ = 2y+3x with y(0) = 1 is y = − 23 x + e2x . 8 Question 9 Question Solve the initial value problem: y ′ + y = e−2x , y(0) = 3 Solution Step 1: First, let’s find the integrating factor µ(x) by multiplying both sides of the differential equation by it: µ(x) = e R 1 dx = ex Step 2: Now, multiply the differential equation by the integrating factor: ex y ′ + ex y = e−x · e−2x Step 3: Notice that the left side can be rewritten as the derivative of the product ex y: (ex y)′ = e−3x Step 4: Integrate both sides with respect to x: Z Z x ′ (e y) dx = e−3x dx Step 5: Solve the integrals: 1 ex y = − e−3x + C 3 Step 6: Next, solve for y by dividing by ex : 1 y = − e−4x + Ce−x 3 Step 7: Use the initial condition y(0) = 3 to find the value of C: 1 3=− +C 3 ⇒ C= 10 3 Step 8: Finally, the solution to the initial value problem is: 1 10 y = − e−4x + e−x 3 3 9 Question 10 Question Consider the initial value problem dy = x2 + y, dx y(0) = 1. Find the particular solution to the initial value problem using the method of integrating factors. Solution Step 1: First, we rewrite the given differential equation in the form y ′ +P (x)y = Q(x) where P (x) and Q(x) are functions of x. dy = x2 + y dx dy − y = x2 . dx Step 2: Next, we identify P (x) and Q(x): Q(x) = x2 . P (x) = −1, e R Step 3: To find the integrating factor I(x), we use the formula I(x) = . P (x) dx I(x) = e R −1 dx = e−x . Step 4: Multiply the integrating factor by the differential equation. e−x dy − e−x y = x2 e−x dx d −x (e y) = x2 e−x . dx Step 5: Integrate both sides with respect to x. Z Z d −x (e y) dx = x2 e−x dx dx e−x y = (−x2 − 2x − 2)e−x + C, where C is the constant of integration. Step 6: Solve for y. y = −x2 − 2x − 2 + Cex . 10 Step 7: To find the particular solution, we substitute x = 0 and y = 1 into the equation. 1 = −02 − 2(0) − 2 + Ce0 1 = −2 + C C = 3. Therefore, the particular solution to the initial value problem is y = −x2 − 2x − 2 + 3ex . Question 11 Question Consider the initial value problem: x2 + y 2 dy = , dx xy y(1) = 2 Use the method of separation of variables to find the particular solution to this initial value problem. Solution Step 1: Rewrite the differential equation by separating variables: x2 + y 2 y x dy = =⇒ 2 dy = 2 dx dx xy y x Step 2: Integrate both sides with respect to their respective variables: Z Z 1 1 dy = dx y x Step 3: Solve the integrals: ln |y| = ln |x| + C where C is the constant of integration. Step 4: Exponentiate both sides to remove the natural logarithm: y = eln |x|+C = eln |x| · eC = Celn |x| = C|x| Step 5: Apply the initial condition y(1) = 2 to find the value of the constant C: 2 = C · 1 =⇒ C = 2 Step 6: Substitute C = 2 back into the general solution to find the particular solution: y = 2|x| Therefore, the particular solution to the initial value problem is y = 2|x|. 11 Question 12 Question Solve the initial value problem: dy + 2y = 4x, dx y(0) = 3 Solution R Step 1: First, we identify the integrating factor, which is given by e 2 dx = e2x . Step 2: Multiply the integrating factor by the differential equation: e2x dy + 2e2x y = 4xe2x dx Step 3: Recognize that the left-hand side is the derivative of the product of y and the integrating factor: d (ye2x ) = 4xe2x dx Step 4: Integrate with respect to x: Z Z d 2x (ye ) dx = 4xe2x dx dx Step 5: Solve the integrals: ye2x = Z 4xe2x dx ye2x = 2xe2x − Z 2e2x dx ye2x = 2xe2x − e2x + C1 Step 6: Solve for y: y = 2x − 1 + C1 e−2x Step 7: Apply the initial condition y(0) = 3 to determine the value of C1 : 3 = 2(0) − 1 + C1 C1 = 4 Step 8: The solution to the initial value problem is: y = 2x − 1 + 4e−2x 12 Question 13 Question Consider the initial value problem dy = 2x(3 − y 2 ), dx y(0) = 1. Find the particular solution to this initial value problem. Solution Step 1: First, let’s separate variables in the given differential equation. dy = 2x dx. 3 − y2 Step 2: Next, we integrate both sides of the equation. Z Z 1 dy = 2 x dx. 3 − y2 Step 3: To integrate the left side, we use partial fraction decomposition. Let A and B be constants such that 1 A B =√ +√ . 2 3−y 3+y 3−y Step 4: Multiplying both sides by 3 − y 2 gives √ √ 1 = A( 3 − y) + B( 3 + y). √ √ Step 5: Set y = 3 to find A and set y = − 3 to find B. Solving the system 1 1 of equations gives A = − 2√ and B = 2√ . 3 3 Step 6: Now, we can rewrite the integral as Z Z Z 1 1 1 1 √ √ − √ dy + √ dy = 2 x dx. 2 3 3+y 2 3 3−y Step 7: After integrating both sides, we get √ √ 1 1 − √ ln | 3 + y| + √ ln | 3 − y| = x2 + C, 2 3 2 3 where C is the constant of integration. Step 8: Simplifying the left side and using the initial condition y(0) = 1, we find C = 0. Step 9: Thus, the particular solution to the initial value problem is √ √ 1 1 − √ ln | 3 + y| + √ ln | 3 − y| = x2 . 2 3 2 3 13 Question 14 Question Solve the initial value problem: dy 2y − x = , dx x+y y(0) = 1. Solution Step 1: Rewrite the differential equation in a more standard form by dividing both sides by x + y: 2y − x dy 2y/x − 1 dy = =⇒ = . dx x+y dx 1 + y/x Step 2: Let u = y/x, then y = ux and the differential equation gives: u+x dy dx = u + x du dx . Substituting this into du 2u − 1 du u−1 = =⇒ x = . dx 1+u dx 1+u Step 3: Separate variables and integrate both sides: Z Z 1+u 1 du = dx. u−1 x Step 4: Solve the integral on the left side: Z Z Z 1+u (u − 1) + 2 2 du = du = 1 + du = u + 2 ln |u − 1| + C. u−1 u−1 u−1 Step 5: Substitute back u = y/x and solve for y: y + 2 ln |y/x − 1| = ln |x| + C. x Step 6: Use the initial condition y(0) = 1 to find the value of C: 1 + 2 ln |1 − 1| = ln |0| + C =⇒ 0 = −∞ + C =⇒ C = ∞. 0 Step 7: The particular solution is then: y + 2 ln |y/x − 1| = ln |x| + ∞. x Therefore, the solution to the initial value problem is ln |x| + ∞. 14 y x + 2 ln |y/x − 1| = Question 15 Question Consider the initial value problem: dy 2x = , dx y y(0) = 1 Solution Step 1: Rewrite the initial value problem as a separable first-order differential equation. y dy = 2x dx Step 2: Integrate both sides of the equation. Z Z y dy = 2x dx Step 3: Solve the integrals. 1 2 y = x2 + C 2 where C is the constant of integration. Step 4: Use the initial condition y(0) = 1 to find the value of the constant C. 1 1 2 · 1 = 02 + C =⇒ C = 2 2 Step 5: Substitute the value of C back into the equation. 1 2 1 y = x2 + 2 2 Step 6: Solve for y in terms of x. y 2 = 2x2 + 1 ⇒ p y = ± 2x2 + 1 √ Therefore, the solution to the initial value problem is y = ± 2x2 + 1, but √ since y(0) = 1, the solution is y = 2x2 + 1. Question 16 Question Solve the initial value problem: 1 dy = sin(2x) + y, dx 2 15 y(0) = 1 Solution Step 1: Rewrite the initial value problem: dy 1 = sin(2x) + y dx 2 with y(0) = 1 Step 2: This is a first-order linear differential equation in the form p(x)y = q(x), where p(x) = 12 and q(x) = sin(2x). R Step 3: We start by finding the integrating factor I(x) = e p(x)dx : R I(x) = e 1 2 dx dy dx + 1 = e2x Step 4: Multiply both sides of the differential equation by the integrating factor I(x): 1 1 1 dy 1 e2x + e 2 x · y = e 2 x sin(2x) dx 2 Step 5: Rewrite the left-hand side as the derivative of the product I(x)y using the product rule: 1 d  1x  e 2 y = e 2 x sin(2x) dx Step 6: Integrate both sides with respect to x: Z Z 1 d  1x  e 2 y dx = e 2 x sin(2x)dx dx Step 7: Integrating the right-hand side with respect to x, we get: 1 1 1 e 2 x y = − e 2 x cos(2x) + C 2 1 Step 8: Solve for y by multiplying through by e− 2 x : 1 1 y = − cos(2x) + Ce− 2 x 2 Step 9: Use the initial condition y(0) = 1 to solve for the constant C: 1 1 1 3 1 = − cos(0) + Ce− 2 ·0 =⇒ 1 = − + C =⇒ C = 2 2 2 Step 10: Substitute the value of C back into the general solution to find the particular solution: 1 3 1 y = − cos(2x) + e− 2 x 2 2 Therefore, the solution to the initial value problem 1 y(0) = 1 is y = − 12 cos(2x) + 32 e− 2 x . 16 dy dx = sin(2x) + 12 y with Question 17 Question Solve the initial value problem: y ′ = 4y − e5x , y(0) = 3 Solution Step 1: First, we find the general solution to the given differential equation. The standard form of the linear first-order differential equation is y ′ + P (x)y = Q(x). Comparing the given equation y ′ = 4y − e5x with the standard form, we have P (x) = 4 and Q(x) = e5x . The integrating factor I(x) is given by: I(x) = e R P (x) dx =e R 4 dx = e4x Multiplying both sides of the differential equation by the integrating factor, we get: e4x y ′ + 4e4x y = e9x Now, we can rewrite the left side of the above equation as the derivative of the product of e4x and y using the product rule: d 4x (e y) = e9x dx Integrating both sides with respect to x gives us: e4x y = 1 9x e +C 9 (Where C is the constant of integration) Step 2: Next, we apply the initial condition y(0) = 3 to find the value of the constant C: 1 e4(0) · 3 = e9(0) + C 9 1 3= +C 9 1 26 C =3− = 9 9 Therefore, the particular solution to the initial value problem is: e4x y = y= 1 9x 26 e + 9 9 1 5x 26 −4x e + e 9 9 17 Question 18 Question Consider the initial value problem given by dy x+y = , dx x−y y(0) = 1. Find the particular solution to the initial value problem. Solution Step 1: Rewrite the initial value problem in a separable form. x+y dy = dx x−y dy x+y = dx x−y (x − y) dy = (x + y) dx Step 2: Integrate both sides to solve for y. Z Z (x − y) dy = (x + y) dx Z Z Z Z x dy − y dy = x dx + y dx x2 y2 y2 = + + C1 2 2 2 where C1 is the constant of integration. Step 3: Use the initial condition y(0) = 1 to find the value of C1 . xy − (0)(1) − (1)2 (0)2 (1)2 = + + C1 2 2 2 1 1 − = + C1 2 2 −1 = 1 + 2C1 C1 = −1 Step 4: Substitute C1 = −1 back into the equation to find the particular solution. y2 x2 y2 xy − = + −1 2 2 2 2xy − 2y 2 = x2 + y 2 − 2 2xy − 2y 2 = x2 + y 2 − 2 2xy + y 2 = x2 + 2 Therefore, the particular solution to the initial value problem is 2xy + y 2 = x + 2. 2 18 Question 19 Question Consider the initial value problem: dy = 2e3x − 5y, dx y(0) = 3 Solution Step 1: First we rewrite the differential equation as a separable equation: dy + 5y = 2e3x dx Step 2: To solve the separable equation, we separate variables: dy dy + 5y = 2e3x =⇒ = 2e3x − 5y dx dx Step 3: We now solve the separable equation: Separating variables, we have: dy = dx 2e3x − 5y Step 4: Next, we integrate both sides of the equation: Z Z 1 dy = dx 2e3x − 5y Step 5: By substituting u = 2e3x − 5y, du = −5dy, we get: Z Z −1 du = dx 5u Step 6: Integrating both sides, we have: −1 ln |u| = x + C 5 Step 7: Substituting back u = 2e3x − 5y and rearranging, we get: ln |2e3x − 5y| = −5x − 5C Step 8: Next, we solve for y: 2e3x − 5y = e−5x−5C Step 9: Applying the initial condition y(0) = 3, we can solve for C: 2e0 − 5(3) = e0 =⇒ 2 − 15 = 1 =⇒ C = −13 Step 10: Substituting this value of C back in, we have: 2e3x − 5y = e−5x−5(−13) Step 11: Simplifying, we get the solution to the initial value problem: 2e3x − 5y = e−5x+65 19 Question 20 Question Solve the initial value problem: dy x = , dx y y(0) = 1 Solution dy = xy . Step 2: Rearrange the Step 1: Rewrite the differential equation as dx terms to separate variables: y dy = x dx. Step 3: Integrate both sides of the equation: Z Z y dy = x dx Step 4: Solve the integrals to get: y2 x2 = +C 2 2 where C is the constant of integration. Step 5: Find the value of the constant C using the initial condition y(0) = 1: 12 02 1 = + C =⇒ =C 2 2 2 So, the equation becomes y2 2 = x2 2 + 21 . Step 6: Solve for y in terms of x: p y 2 = x2 + 1 =⇒ y = ± x2 + 1 √ Step 7: Since y(0) = 1, we choose y = x2 + 1. Therefore, the solution to the √ initial value problem is y = x2 + 1. Question 21 Question Consider the initial value problem: dy = 2y + x, dx y(0) = 1 Find the particular solution to the differential equation that satisfies the initial condition. 20 Solution dy Step 1: First, let’s solve the differential equation dx = 2y +R x using the method of integrating factors. The integrating factor is given by e 2dx = e2x . d 2x (e y) = e2x (2y + x) dx Z e2x y = e2x (2y + x)dx Z Z e2x y = 2e2x ydx + xe2x dx Step 2: We can solve the integrals on the right side of the equation to get: Z Z 2x 2x e y = 2 e ydx + xe2x dx Z Z 1 1 e2x dx e2x y = 2e2x y − e2x dy + xe2x − 2 2 1 1 e2x y = 2e2x y − e2x + xe2x − e2x + C 2 4 Step 3: Now, applying the initial condition y(0) = 1, we get: e0 · 1 = 2e0 · 1 − e0 + 1=2−1− 1 1 · 0 · e0 − · e0 + C 2 4 1 4 1 +C 4 3 C= 4 1= Step 4: Substituting C = 3 4 back into our equation, we find: 1 1 3 e2x y = 2e2x y − e2x + xe2x − e2x + 2 4 4 1 2x 1 2x 3 2x 2x 2x e y − 2e y = −e + xe − e + 2 4 4 1 2x 1 2x 3 2x 2x −e y = −e + xe − e + 2 4 4 1 1 3 −2x y =1− x+ + e 2 4 4 Therefore, the particular solution to the differential equation that satisfies the initial condition y(0) = 1 is y = 1 − 12 x + 14 + 34 e−2x . 21 Question 22 Question Consider the initial value problem: dy = 2x + 3y, dx y(0) = 1. Find the particular solution to the initial value problem. Solution Step 1: Rewrite the given initial value problem as: dy − 3y = 2x, dx y(0) = 1. dy Step 2: This is a first-order linear differential Requation of the form dx + P (x)dx P (x)y = Q(x). The integrating factor is given by e . R Step 3: In this case, P (x) = −3, so the integrating factor is e −3dx = e−3x . Step 4: Multiply both sides of the differential equation by the integrating factor: dy − 3e−3x y = 2xe−3x . e−3x dx Step 5: Rewrite the equation as the derivative of a product: d (ye−3x ) = 2xe−3x . dx Step 6: Integrate both sides with respect to x: Z Z d (ye−3x ) dx = 2xe−3x dx. dx Step 7: This simplifies to: 2 2 ye−3x = − xe−3x − e−3x + C, 3 9 where C is the constant of integration. Step 8: Applying the initial condition y(0) = 1, we have: 1=0− 2 +C 9 ⇒ C= 11 . 9 Step 9: Substitute C = 11 9 back into the equation to find the particular solution: 2 11 2 ye−3x = − xe−3x − e−3x + . 3 9 9 Thus, the particular solution to the initial value problem is y = − 23 x − 92 + 11 3x 9 e . 22 Question 23 Question Solve the initial value problem: y ′ = 3y − 6, y(0) = 2 Solution Step 1: First, let’s solve the differential equation y ′ = 3y − 6 by separating variables: dy = 3y − 6 dt dy = dt 3y − 6 Step 2: Integrate both sides with respect to y and t: Z Z 1 dy = dt 3y − 6 1 ln |3y − 6| = t + C1 3 ln |3y − 6| = 3t + C2 |3y − 6| = e3t+C2 |3y − 6| = Ce3t 3y − 6 = Ce3t (where C = eC2 ) 3y = Ce3t + 6
1 Ce3t + 6 y= 3 Step 3: Next, we apply the initial condition y(0) = 2 to find the value of C: 2= or
1 C · e0 + 6 3 1 (C + 6) 3 6=C +6 2= C=0 Step 4: Now that we have C = 0, the solution to the initial value problem is: y= 1 · 6 = 2e3t 3 23 Question 24 Question Solve the initial value problem: dy 3x2 = , dx y y(1) = 2 Solution Step 1: Rewrite the differential equation in a separable form. y dy = 3x2 dx Step 2: Integrate both sides with respect to their respective variables. Z Z y dy = 3x2 dx Step 3: Solve the integrals. 1 2 y = x3 + C 2 where C is the constant of integration. Step 4: Apply the initial condition y(1) = 2 to find the value of C. 1 2 (2) = 13 + C =⇒ 2 = 1 + C =⇒ C = 1 2 Step 5: Substitute the value of C back into the equation. 1 2 y = x3 + 1 2 Step 6: Solve for y in terms of x. y 2 = 2x3 + 2 =⇒ y = ± p 2x3 + 2 Therefore, the solution to the initial value problem is y = y(1) = 2. Question 25 Question Solve the initial value problem: dy = 2x + y, dx 24 y(0) = 1 √ 2x3 + 2 since Solution Step 1: Write the given initial value problem as: dy − y = 2x, y(0) = 1 dx Step 2: This is a first-order linear differential equation.R To solve it, we first find the integrating factor µ(x) which is given by µ(x) = e −1 dx . µ(x) = e R −1 dx = e−x Step 3: Multiply both sides of the differential equation by the integrating factor µ(x): dy − e−x y = 2xe−x e−x dx Step 4: Recognize the left side as the derivative of the product of e−x and y (this step involves integrating factors): d (ye−x ) = 2xe−x dx Step 5: Integrate both sides with respect to x: Z Z d −x (ye ) dx = 2xe−x dx dx Step 6: Integrate to get: ye−x = −2xe−x − 2e−x + C Step 7: Solve for y by multiplying both sides by ex : y = −2x − 2 + Cex Step 8: Use the initial condition y(0) = 1 to find C: 1 = −2(0) − 2 + C C=3 Step 9: Substitute C = 3 back into the general solution to get the particular solution: y = −2x − 2 + 3ex Therefore, the solution to the initial value problem is y = −2x − 2 + 3ex with the initial condition y(0) = 1. Question 26 Question Solve the initial value problem: y ′ = 2xy 2 , 25 y(0) = 1. Solution Step 1: Let’s rewrite the differential equation in a form suitable for separation of variables: dy = 2xy 2 . dx Step 2: Separate variables by dividing both sides by y 2 and multiplying by dx: dy = 2xdx. y2 Step 3: Integrate both sides: Z 1 dy = 2 y2 Z xdx. Step 4: Solving the integrals gives: − 1 = x2 + C, y where C is the constant of integration. Step 5: To find the value of the constant C, we use the initial condition y(0) = 1: 1 − = 02 + C ⇒ C = −1. 1 Step 6: Substituting C back into the equation gives: − 1 = x2 − 1. y Step 7: Solve for y: y= 1 . 1 − x2 Therefore, the solution to the initial value problem is y = initial condition y(0) = 1. Question 27 Question Solve the initial value problem: dy 2x = 2, dx y 26 y(0) = 1 1 1−x2 with the Solution Step 1: Separate variables by writing the given differential equation as: y 2 dy = 2xdx Step 2: Integrate both sides with respect to their respective variables: Z Z y 2 dy = 2xdx Step 3: Integrate to solve for y: y3 = x2 + C 3 where C is the constant of integration. Step 4: Applying the initial condition y(0) = 1 to solve for C: 1 13 = 0 + C =⇒ C = 3 3 Step 5: Substitute the value of C back into the equation to find the particular solution: y3 1 = x2 + =⇒ y 3 = 3x2 + 1 3 3 Step 6: Solve for y by taking the cube root of both sides: p 3 y = 3x2 + 1 √ Therefore, the solution to the initial value problem is y = 3 3x2 + 1 with the initial condition y(0) = 1. Question 28 Question Consider the initial value problem: 2x dy = , dx y y(0) = 2. Solution Step 1: Rewrite the differential equation in a separable form by isolating variables x and y: y dy = 2x dx. Step 2: Integrate both sides of the equation: Z Z y dy = 2x dx. 27 Step 3: Solve the integrals: y2 = x2 + C, 2 where C is the constant of integration. Step 4: Use the initial condition y(0) = 2 to find the value of the constant C: 22 = 0 + C =⇒ C = 2. 2 Step 5: Substitute the value of C back into the equation: y2 = x2 + 2. 2 Step 6: Solve for y in terms of x: p y 2 = 2(x2 + 2) =⇒ y = ± 2x2 + 4. Therefore, the solution to the initial value problem is y = y(0) = 2. √ 2x2 + 4 since Question 29 Question Solve the initial value problem: dy 4x + 3y − 6 = , dx 3x + 2y − 5 y(0) = 2 Solution Step 1: First, we rewrite the given differential equation in terms of dy and dx by multiplying both sides by dx: dx · dy 4x + 3y − 6 = · dx dx 3x + 2y − 5 Step 2: To make it easier to solve, we rewrite the equation as follows: 3x + 2y − 5 dy = dx 4x + 3y − 6 Step 3: We now integrate both sides, and find a constant of integration, C: Z Z 3x + 2y − 5 dy = dx 4x + 3y − 6 28 Step 4: After integrating, we get: 1 ln |4x + 3y − 6| = x + C 2 Step 5: We can rewrite the expression using exponential functions: |4x + 3y − 6| = e2x+2C Step 6: Since y(0) = 2, we substitute x = 0 and y = 2 into the equation: |4(0) + 3(2) − 6| = e2(0)+2C Step 7: Simplifying, we get: |6| = e2C Step 8: Thus, 6 = e2C or e2C = 6. Taking the natural logarithm of both sides, we get: 2C = ln 6 Step 9: Solving for C, we have: C= 1 ln 6 2 Step 10: Substitute the value of C back into the equation we obtained: |4x + 3y − 6| = e2x+ln 6 Step 11: Finally, we simplify the equation to get the solution to the initial value problem: 4x + 3y − 6 = 6e2x Question 30 Question Consider the initial value problem: dy = 2x3 + 3x2 + x − 1, dx y(0) = 5 Use Euler’s method with step size h = 0.1 to approximate the value of y(0.2). 29 Solution Step 1: Define the step size h, the initial condition, and the function f (x, y). ˆ Step size: h = 0.1 ˆ Initial condition: y(0) = 5 ˆ Function f (x, y) = 2x3 + 3x2 + x − 1 Step 2: Perform Euler’s method to approximate y(0.1), y(0.2), . . . until x = 0.2. x1 = x0 + h = 0 + 0.1 = 0.1 y1 = y0 + hf (x0 , y0 ) = 5 + 0.1(2(0)3 + 3(0)2 + 0 − 1) = 5 − 0.1 = 4.9 x2 = x1 + h = 0.1 + 0.1 = 0.2 y2 = y1 + hf (x1 , y1 ) = 4.9 + 0.1(2(0.1)3 + 3(0.1)2 + 0.1 − 1) = 4.9 + 0.1(0.002 + 0.03 + 0.1 − 1) = 4.9 + 0.1(−0.968) = 4.9 − 0.0968 = 4.8032 Therefore, the approximate value of y(0.2) using Euler’s method with step size h = 0.1 is y(0.2) ≈ 4.8032. Question 31 Question Solve the initial value problem: 2x dy = , dx y y(0) = 1 Solution Step 1: Rewrite the differential equation in terms of y and dy: y dy = 2x dx Step 2: Integrate both sides: Z Z y dy = 2x dx y2 = x2 + C1 2 30 Step 3: Apply the initial condition y(0) = 1: 12 = 0 + C1 2 1 C1 = 2 Step 4: Substitute C1 = 1 2 back into the equation: y2 1 = x2 + 2 2 y 2 = 2x2 + 1 Step 5: Solve for y by taking the square root: p y = ± 2x2 + 1 Step 6: Apply the initial condition y(0) = 1 to determine the specific solution: p 1 = 2(0)2 + 1 √ 1= 1 y=1 Therefore, the solution to the initial value problem is y = 1. Question 32 Question Solve the initial value problem: dy dx = y 2 + x2 , y(0) = 1. Solution Step 1: Rewrite the equation as a separable first-order differential equation by bringing all terms involving y to the left side and all terms involving x to the right side. dy = y 2 + x2 dx dy = dx y2 Step 2: Integrate both sides with respect to y and x separately. Z Z 1 dy = dx y2 − 1 = x + C1 y 31 where C1 is the constant of integration. Step 3: Solve for y in terms of x by isolating y. − 1 = x + C1 y −1 = y(x + C1 ) y= 1 1 − x − C1 Step 4: Use the initial condition y(0) = 1 to find the value of the constant C1 . 1 1 − 0 − C1 1 1= 1 − C1 1= 1 − C1 = 1 C1 = 0 Step 5: Substitute C1 = 0 back into the expression for y to obtain the particular solution. 1 y= 1−x Therefore, the solution to the initial value problem is y = 1 1−x . Question 33 Question Solve the initial value problem: dy = 2x(1 − y), dx y(0) = 0 Solution Step 1: First, rewrite the differential equation in a separable form by moving all terms involving y to one side and all terms involving x to the other side: dy = 2x dx 1−y Step 2: Integrate both sides with respect to their respective variables: Z Z 1 dy = 2x dx 1−y 32 Step 3: Solve the integrals: − ln |1 − y| = x2 + C1 where C1 is the constant of integration. Step 4: Exponentiate both sides to eliminate the natural logarithm: |1 − y| = e−x 2 −C1 = Ce−x 2 where C = ±e−C1 . Step 5: Rewrite the equation without absolute values: 1 − y = Ce−x 2 y = 1 − Ce−x 2 or Step 6: Apply the initial condition y(0) = 0 to solve for the constant C: 0 = 1 − Ce−0 C=1 Step 7: Substitute C = 1 back into the solution: y = 1 − e−x 2 2 Therefore, the solution to the initial value problem is y = 1 − e−x . Question 34 Question Solve the initial value problem: 1 dy = , dx x+y y(1) = 1 Solution Step 1: Rewrite the differential equation in standard form by separating variables: (1 + y) dy = dx Step 2: Integrate both sides with respect to their respective variables: Z Z (1 + y) dy = dx Z Z 1 dy + Z y dy = 33 dx y2 =x+C 2 Step 3: Use the initial condition y(1) = 1 to find the value of the constant y+ C: 1+ Step 4: Substitute C = solution: 1 2 1 =1+C 2 1 C= 2 back into the equation to find the particular y2 1 =x+ 2 2 Step 5: Use the quadratic formula to solve for y: y+ y 2 + 2y − 2x − 1 = 0 p −2 ± 22 − 4 · 1 · (−2x − 1) y= 2·1 √ −2 ± 4 + 8x + 4 y= 2 √ −2 ± 8x + 8 y= 2 √ y = −1 ± 2x + 2 Step 6: Use the initial condition y(1) = 1 to determine the specific solution: √ 1 = −1 ± 2 · 1 + 2 √ 2= 4 √ Since 4 = 2 > 1, we choose the positive sign: √ y = −1 + 2x + 2 √ Therefore, the solution to the initial value problem is y = −1 + 2x + 2. Question 35 Question Consider the initial value problem: dy = 2y, dx y(0) = 3 Find the solution to this initial value problem. 34 Solution Step 1: We start by separating variables in the differential equation. dy dy = 2y ⇒ = 2dx dx y Step 2: Next, we integrate both sides. Z Z 1 dy = 2dx ⇒ ln |y| = 2x + C y Step 3: Apply the initial condition y(0) = 3 to solve for the constant C. ln |3| = 2 · 0 + C ⇒ ln |3| = C Step 4: Substitute back the value of C into the equation obtained in Step 2. ln |y| = 2x + ln |3| Step 5: Solve for y by removing the natural logarithm. |y| = e2x+ln |3| = e2x · 3 Step 6: Consider both the positive and negative cases for y. y = ±3e2x Therefore, the solution to the initial value problem is y = 3e2x or y = −3e2x . 35