Differential Equations Homework Solutions 2.1-2.2 plus Substitutions

Homework Solutions: 2.1-2.2, plus Substitutions Section 2.1 I have not included any drawings/direction fields. We can see them using Maple or by hand, so we’ll be focusing on getting the analytic solutions here: 1. Problem 1: Solve the DE using the Method of Integrating Factor (for linear differential equations): y 0 + 3y = t + e−2t ⇒  e3t (y 0 + 3y) = e3t t + e−2t  ⇒  0 e3t y(t) = te3t + et Integrate both sides Hint: We need to use “integration by parts” to integrate te3t . Using a table as in class: + t e3t − 1 (1/3)e3t + 0 (1/9)e3t Z ⇒ 1 1 te3t dt = e3t − e3t 3 9 Putting it all together, 1 1 e3t y(t) = te3t − e3t + et + C 3 9 so that 1 1 1 C y(t) = t − + −2t + 3t 3 9 e e Notice that the last two terms go to zero as t → ∞, so we see that y(t) does approach a line: 1 1 t− 3 9 as t → ∞. 2. Problem 3: Very similar situation to Problem 1. Let’s go ahead and solve: y 0 + y = te−t + 1 Multiply both sides by e R p(t) dt = et : et (y 0 + y) = t + et ⇒  0 et y(t) = t + et Integrate both sides: 1 et y(t) = t2 + et + C 2 ⇒ This could be written as: y(t) = 1 + so that it is clear that, as t → ∞, y(t) → 1. 1 y(t) = t2 e−t + 1 + Ce−t 2 t2 C + t t 2e e 3. Problem 5: y 0 − 2y = 3et  ⇒ e−2t y 0 = e−t Finishing, we get: y(t) = −3et + Ce2t (All solutions tend to ±∞) 4. Problem 7: 2 y 0 + 2ty = 2tet  ⇒ 2 et y 0 = 2t 2 so that y(t) = (t2 + C)e−t , and all solutions tend to zero as t → ∞. 5. Problem 13: (You’ll need to integrate by parts!) y 0 − y = 2te2t e R p(t) dt = e−t y(t) = e2t (2t − 2) + 3et 6. Problem 15: ty 0 + 2y = t2 − t + 1 Be sure to put in standard form before solving: 2 1 y0 + y = t − 1 + t t and e R p(t) dt = t2 1 1 1 1 y(t) = t2 − t + + 4 3 2 12t2 7. Problem 16: In this problem, the integrating factor is again t2 : y0 + 2 cos(t) ·y = t t2 ⇒ y(t) = sin(t) t2 8. Problem 30: y 0 − y = 1 + 3 sin(t) y(0) = y0 This is very similar to Problem 29. Note that: Z 1 e−t sin(t) dt = − e−t (cos(t) + sin(t)) 2 Therefore, the general solution is (details left out): 3 5 + y 0 et y(t) = −1 − (cos(t) + sin(t)) + 2 2   To keep the solution finite (or bounded) as t → ∞, we must find y0 so that the exponential term drops out- This means that y0 = −5/2. 9. Problem 35: There are many ways of constructing such a differential equation- It’s easiest to start with a desired solution. We’ll again show two possibilities: • If we would like y(t) = 3 − t + Ce−3t , then y 0 = −1 − 3Ce−3t , and: y 0 + 3y = 8 − 3t • If we would like y(t) = 3 − t + Ct , then y 0 = −1 − C/t2 , and we see that: ty 0 + y = 3 − 2t 10. Problem 36: Similar to 35, let’s try a solution then construct a DE. In this case, if y = 2t − 5 + Ce−t , then y 0 = 2 − Ce−t , so that: y 0 + y = 2t − 3 2.2 1. Problem 1: Give the general solution: y 0 = x2 /y y dy = x2 dx 1 2 1 3 y = x +C 2 3 ⇒ 2. Problem 3: Give the general solution to y 0 + y 2 sin(x) = 0. First write in standard form: dy = −y 2 sin(x) dx ⇒ − 1 dy = sin(x) dx y2 Before going any further, notice that we have divided by y, so we need to say that this is value as long as y(x) 6= 0. In fact, we see that the function y(x) = 0 IS a possible solution. With that restriction in mind, we proceed by integrating both sides to get: 1 = − cos(x) + C y ⇒ y= 1 C − cos(x) 3. Problem 5: This one is good for a little extra practice in integrating. Recall that: Z cos2 (x) dx = 1Z 1 1 (1 + cos(2x)) dx = x + sin(2x) 2 2 2   Given that, y 0 = cos2 (x) cos2 (2y) ⇒ sec2 (2y) dy = cos2 (x) dx ⇒ 1 1 1 tan(2y) = x + sin(2x) + C 2 2 2 It would be fine to leave it in this form. Of course, this solution is only valid when cos(2y) 6= 0. The constant solutions   cos(2y) = 0 would also be solutions (we’ll focus on these types of solutions later). Solving for y, we get the answer in the text. 4. Problem 7: Give the general solution: dy x − e−x = dx y + ey First, note that dy/dx exists as long as y 6= ey . With that requirement, we can proceed:   (y + ey ) dy = x + e−x dx Integrating, we get: 1 1 2 y + ey = x2 − e−x + C 2 2 In this case, we cannot algebraically isolate y, so we’ll leave our answer in this form (we could multiply by two). 5. Problem 9: Let y 0 = (1 − 2x)y 2 , y(0) = −1/6. First, we find the solution. Before we divide by y, we should make the note that y 6= 0. We also see that y(x) = 0 is a possible solution (although NOT a solution that satisfies the initial condition). Now solve: Z y −2 dy = Z (1 − 2x) dx ⇒ −y −1 = x − x2 + C Solve for the initial value: 6=0+C ⇒C =6 The solution is (solve for y): y(x) = x2 1 1 = −x−6 (x − 3)(x + 2) The solution is valid only on −2 < x < 3, and we could plot this by hand (also see the Maple worksheet). 6. Problem 11: x dx + ye−x dy = 0, y(0) = 1 To solve, first get into a standard form, multiplying by ex , and integrate (integration by parts for the right hand side): Z y dy = − Z xex dx ⇒ 1 2 y = −xex + ex + C 2 We could solve for the constant before isolating y: 1 =0+1+C 2 C=− 1 2 Now solve for y: y 2 = 2ex (x − 1) − 1 2 and take the positive root, since y(0) = +1. y= q 2ex (1 − x) − 1 The solution exists as long as: 2ex (1 − x) − 1 ≥ 0 We would have to use Maple to solve where this is equal to zero- Given software, we could see that −1.678 ≤ x ≤ 0.768. 7. Problem 16: x(x2 + 1) 1 dy √ = y(0) = − dx 4y 3 2 First, we notice that y 6= 0. Now separate the variables and integrate: 1 1 y 4 = x4 + x2 + C 4 2 This might be a good time to solve for C: C = 1/4, so: 1 1 1 y 4 = x4 + x2 + 4 2 4 The right side of the equation seems to be a nice form. Try some algebra to simplify it:  1 4 1 x + 2x2 + 1 = (x2 + 1)2 4 4 Now we can write the solution: 1 1 √ y 4 = (x2 + 1)2 ⇒ y = − √ x2 + 1 4 2 This solution exists for all x (it is the bottom half of a hyperbola- see the Maple plot). √ 8. Problem 20: y 2 1 − x2 dy = sin1 (x) dx with y(0) = 1. To put into standard form, we’ll be dividing so that x 6= ±1. In that case, Z 2 y dy = sin−1 (x) √ dx 1 − x2 Z −1 The right √ side of the equation is all set up for a u, du substitution, with u = sin (x), du = 1/ x2 − 1 dx: 1 3 1 y = (arcsin(x))2 + C 3 2 Solve for C, 31 = 0 + C so that: 1 3 1 1 y = arcsin2 (x) + 3 2 3 Now, s 3 arcsin2 (x) + 1 2 The domain of the inverse sine is: −1 ≤ x ≤ 1. However, we needed to exclude the endpoints. Therefore, the domain is: y(x) = 3 −1 < x < 1 Substitution Methods See the PDF file that is linked online - It shows the direction fields so that you can see the “zoom invariance” we were talking about in class. • pg. 50, 31: y 0 = (x2 + xy + y 2 )/x2 Divide it out:  2 y y y =1+ + x x 0 Our substitution will be v = y/x, or y = xv so that y 0 = v + xv 0 : v + xv 0 = 1 + v + v 2 ⇒ xv 0 = 1 + v 2 This is now separable: 1 1 dv = dx 2 1+v x ⇒ tan−1 (v) = ln |x| + C ⇒ Remember to back-substitute. You can typically leave your answer in implicit form: tan−1 y = ln |x| + C x   • p. 50, 33: Divide numerator and denominator by x and substitute v = y/x, y = xv and y 0 = v + xv 0 to translate the differential equation to: v + xv 0 = 4v − 3 2−v xv 0 = ⇒ 4v − 3 v(2 − v) v 2 + 2v − 3 − = 2−v 2−v −(v − 2) Separation of variables gives: Z Z −(v − 2) 1 dv = dx 2 v + 2v − 3 x To integrate the left side, we use partial fractions: −v + 2 A B −5 1 1 1 = + = + · v 2 + 2v − 3 v+3 v−1 4 v+3 4 v−1 Now integrate through, and multiply by 4: 5 1 − ln |v + 3| + ln |v − 1| = ln |x| + C 4 4 Back-substitute. We can simplify a bit: ln v−1 = ln(x4 ) + C (v + 3)4 ⇒ v−1 = Ax4 (v + 3)5 ⇒ 5 y y − 1 = Ax4 + 3 x x Multiply both sides by |x| to get the answer in the text, |y − x| = A|y + 3x|5 Of course, this was only valid if v − 1 6= 0 and v + 3 6= 0. Notice that in these cases, we do have solutions: y=x y = −3x (Try substituting them back in!) • p. 50, 35: y 0 = (x + 3y)/(x − y) Substitute as usual, then simplify: v + xv 0 = 1 + 3v 1−v ⇒ xv 0 = (v + 1)2 −v + 1 Separate variables, and integrate (You’ll need partial fractions, shown below): Z where −v + 1 dv = ln |x| + C (v + 1)2 −v + 1 A B 1 2 = + = − + (v + 1)2 v + 1 (v + 1)2 v + 1 (v + 1)2 And this antiderivative is − ln |v + 1| − 2 . v+1 − ln |v + 1| − Put it together to get: 2 = ln |x| + C v+1 Back substitute for v and simplify: ln x 2x − = ln |x| + C y+x y+x ⇒ ln |y + x| + 2x = C2 y+x (NOTE: We do want to simplify some; I’m showing how to get the form in the back of the book) • p. 77, 28: For the Bernoulli equations, we want to divide everything by the highest power of y. In this case (We also divide by t2 to get standard form): y0 2 1 1 + = 2 3 2 y ty t Now if v = 1/y 2 , then dv dy = −2y −3 dx dx y0 1 0 = − v y3 2 ⇒ Substitute back in: 1 2 1 − v0 + v = 2 2 t t Now put this into a standard linear form, then get the integrating factor: 4 2 v0 − v = − 2 t t ⇒ e R p(t) dt = t−4 Therefore, v 0 2 2 = − 6 ⇒ v = 5 + Ct4 4 t t 5t Back substitute and make the right side of the equation into a single fraction for later:   1 2 + C2 t5 = y2 5t which is where the answer in the text comes from. • p. 77, 29: Similarly, y 0 = ry − ky 2 ⇒ y 0 − ry = −ky 2 ⇒ y0 1 − r = −k 2 y y From which we get v= 1 y ⇒ y0 = −v 0 y2 so that −v 0 − rv = −k ⇒ v 0 + rv = k ⇒ From which we get: k + Ce−rt r Back substitute v = 1/y to get the answer in the text. v= (vert )0 = kert