Differential Equations Section 1.3 - Study Guide

Section 1.3 1. Problem 1: Order is 2, and it is linear (divide by the leading t2 ) 2. Problem 3: Order is 4, and it is linear. 3. Problem 5: Order is 2, and nonlinear (because of sin(t + y) term). 4. Problem 7: Do you know the definition of cosh(t)? See our class website before doing this problem- There are practice problems there). You might use the definition directly, or from the practice sheet, see that: d (cosh(x)) = sinh(x) dx d (sinh(x)) = cosh(x) dx Now, to solve problem 7, we want to verify that either y(t) = et or y(t) = cosh(t) satisfies the differential equation: y 00 − y = 0. If y(t) = et , then y 0 (t) = et , and y 00 (t) = et , so y 00 − y = et − et = 0 If y(t) = cosh(t), then y 0 = sinh(t) and y 00 (t) = cosh(t), so again, y 00 − y = cosh(t) − cosh(t) = 0 5. Problem 9: Show that y(t) = 3t + t2 satisfies the ODE: ty 0 − y = t2 . First compute the derivative, then substitute into the expression: y 0 = 3 + 2t so that: ty 0 − y = t(3 + 2t) − (3t + t2 ) = 3t + 2t2 − 3t − t2 = t2 6. Problem 14: Show that the function 2 y(t) = et Z t 2 2 e−s ds + et 0 solves: y 0 − 2ty = 1. To show this directly, we need to recall how to differentiate a function like: g(t) = Z t f (s) ds 0 From the Fundamental Theorem of Calculus, g 0 (t) = f (t). Therefore, if y(t) is as given above, the derivative is found by using the product rule: 0  y = 2te t2  Z t · 2 2 2 2 e−s ds + et e−t + 2tet 0 If we simplify a bit, and subtract: 0 y = 2te t2 Z t 2 2 e−s ds + 1 + 2tet 0  −2ty = −2t e t2 Z t e −s2 t2  ds + e 0 We see that the only remaining term is 1. (NOTE: In Section 2.1, we’ll see where this strange integral is coming from) 7. Problem 15: We did something similar in class: If y = ert , substitute it into the differential equationy 0 + 2y = 0 ⇒ rert + 2ert = 0 Now solve for r: (r + 2)ert = 0 ⇒ ⇒ r+2=0 r = −2 Note that ert = 0 has no solution. Conclusion: y(t) = e−2t . Side Remark: We solved this in Section 1.2 by doing this: y 0 = −2y ⇒ 1 dy = −2 dt y Z ⇒ Z 1 dy = −2 dt y so that: ln |y| = −2t + c ⇒ y(t) = Ae−2t 8. Same setup as Problem 15: If y(t) = ert , y 0 (t) = rert y 00 (t) = r2 ert Substitute these into the DE: y 00 − y 0 − 6y = 0 and solve for r: r2 ert + rert − 6ert = 0 ⇒   ert r2 + r − 6 = 0 Again, ert = 0 has no solution, so just solve: r2 + r − 6 = 0 (r + 3)(r − 2) = 0 Either y = e−3t or y = e2t will solve the DE. r = −3, 2 9. Problem 19: In this case, assume y = tr , so y 0 = rtr−1 and y 00 = r(r −1)tr−2 . Substitute these into the DE: t2 y 00 + 4ty 0 + 2y = 0 t2 · r(r − 1)tr−2 + 4t · rtr−1 + 2tr = 0 ⇒ Simplify and factor out tr : tr (r(r − 1) + 4r + 2) = 0 This equation must be true for ALL t > 0 (given in the problem), so tr = 0 does not give a solution. Solve for r: r2 − r + 4r + 2 = 0 ⇒ 1 t 1 t2 Therefore, y(t) = and y(t) = r2 + 3r + 2 = 0 ⇒ (r + 1)(r + 2) = 0 solve the differential equation. 10. Problem 21: The order is 2, linear. 11. Problem 25: Show that each of these: u(x, y) = ln(x2 + y 2 ) u(x, y) = cos(x) cosh(y) solve the Partial Differential Equation (PDE): uxx + uyy = 0 If u(x, y) = cos(x) cosh(y), then ux = − sin(x) cosh(y) uxx = − cos(x) cosh(y) Similarly, uy = cos(x) sinh(y) uyy = cos(x) cosh(y) And if we add uxx to uyy , we get zero. If u(x, y) = ln(x2 + y 2 ), then: ux = 2x x2 + y 2 uxx = −2(x2 − y 2 ) (x2 + y 2 )2 uy = 2y 2 x + y2 uyy = −2(y 2 − x2 ) (x2 + y 2 )2 Similarly, And again we see that if we add uxx and uyy , we get zero.