we get y 2 5y = x + e x + C: From the initial condition y(0) = 1, we get 1 5 = 0+1+C; so that C = 5. Completing the square to solve y 2 5y = x + e x 5

Transcription

1 Math 24 Final Exam Solution 17 December Find the general solution to the differential equation ty 0 +2y = sin t. Solution: Rewriting the equation in the form (for t 6= 0),we find that y t y = sin t t exp(2 ln jtj) =t 2 is an integrating factor. Multiplying through, we get t 2 y 0 +2ty = t sin t: (1) We recognize the left-hand side of (1) as the derivative oft 2 y,andso we have Integrating by parts, we get t 2 y = Z t sin tdt: t 2 y = t cos t +sint + C; so the general solution to the given equation is 2. Solve the initial value problem y = cos t t Express your answer as a function of x. + sin t t 2 + C t 2 : dy dx = x 2 + e x ; y(0) = 1: 2y 5 Solution: The differential equation is separable; integrating both sides of 2y 5 dy = x 2 + e x dx;

2 we get y 2 5y = x + e x + C: From the initial condition y(0) = 1, we get 1 5 = 0+1+C; so that C = 5. Completing the square to solve y 2 5y = x + e x 5 for y, we have so that y = x + e x y = 5 2 ± sx + e x : Using the initial condition y(0) + 1, once again, we get 1 = 5 2 ± s 9 4 ; telling us that we need the negative square root. The solution is y = 5 2 sx 2 + e x :. Consider the initial value problem y 00 4y 0 +1y =0; y(0) = 2; y 0 (0) = m where m is some constant. a. Find the solution y(t) to the given problem. Your solution will depend on m.

3 b. Suppose the solution to this problem is written in the form y(t) =e at cos(bt + '). Find the value of m for which ' =0. (Hint: Once you've done part (a), this requires almost no computation.) Solution: a. The roots of the characteristic equation r 2 4r +1are 2 ± i, so the general solution to the differential equation is From this, we get y(t) = e 2t (c 1 cos t + c 2 sin t): y 0 (t) = 2e 2t (c 1 cos t + c 2 sin t)+e 2t (c 2 cos t c 1 sin t): Using the initial conditions, we get and 2 = y(0) = c 1 m = y 0 (0) = 2c 1 +c 2 = 4+c 2 : Thus c 1 =2and c 2 = m 4. The solution to the IVP is y(t) = e 2t 2 cos t + m 4 sin t b. Our solution has the form e at cos(bt +0)=e at cos(bt) when the coefficient of the sine term is zero. This is the case only when m =4. 4. Consider the differential equation y 00 +4y = g(t). a. For each forcing function g(t) given below, write down the form of a particular solution y p to the differential equation. Do not try to determine the coefficients. i. g(t) =t(t + 1). :

4 ii. g(t) =(t + 1) sin 2t. iii. g(t) =e t sin 2t. b. Find a forcing function g(t) for which the system exhibits resonance. Solution: We first observe that the homogeneous solution to the given equation is c 1 sin 2t + c 2 cos 2t: a. i. For a polynomial of degree 2, we take y p = At 2 + Bt + C: ii. For a first-degree polynomial times sine, we would ordinarily need the terms sin 2t, cos2t, t sin 2t, andt cos 2t. However, sin 2t and cos 2t are in the homogeneous solution, so we need to multiply everything by t, and our guess takes the form y p = (At 2 + Bt)sin2t +(Ct 2 + Dt) cos 2t: iii. Nothing here appears in the homogeneous solution, so the answer is simply y p = Ae t sin 2t + Be t cos 2t: b. Since the terms sin 2t and cos 2t are in the homogeneous solution, if the forcing function has the form A cos 2t + B sin 2t where either A or B is non-zero, there will be terms in the solution of the form t sin2t or t cos2t. This is the resonance" phenomenon. 5. a. The differential equation y 0 = y y 2 +2y has three equilibrium solutions. Find them, and classify each as stable, unstable, or semi-stable. b. Find positive value k for which the differential equation y 0 = y y 2 +2y + k has exactly two equilibrium solutions. Solution: a. Let f(y) =y y 2 +2y, and note that f(y) =y(y 1)(y 2). The roots of f(y) are the equilibrium solutions of y 0 = f(y). That is, we have equilibrium solutions at y =0,y =1,andy =2. A sketch of f versus y shows that y 0 is negative for y < 0 and 1 <y< 2, and y 0 is positive for 0 <y<1andy>2.

5 Thus the equilibrium solution y = 0 is unstable; the equilibrium solution y = 1 is (asymptotically) stable; and the equilibrium solution y = 2 is unstable. b. The constant k needs to be the absolute value of f at its local minimum between y =1and y =2. To find this value, we solve 0 = f 0 (y) = y 2 6y +2: p By the quadratic formula, the critical numbers of f are 1 ±. The local minimum occurs at y =1+ p. Since f(y) = y(y 1)(y 2), we may calculate the value of f at this critical number as ψ p! ψp!ψp!ψp! f 1+ = ψp! = = 2p 9 : Thus when k = 2p 9 there are exactly two equilibrium solutions. 6. a. Find L n te 2t o. b. Find L n u 2 (t)e 2t o.

6 c. Let y be a solution to the initial value problem y 00 +5y 0 +6y = g(t); y(0) = y 0 (0) = 0 where g is of exponential order as t!1(so that its Laplace transform exists). Find y(t). Your answer will be an integral involving g. Solution: a. From line 10 on the transforms table, we get L n te 2t o = 1 (s +2) 2 : b. We have L n u 2 (t)e 2t o = L n u 2 (t)e 2t+4 e 4 o = e 4 L n u 2 (t)e 2(t 2) o = e 4 e 2s F (s) where F (s) isthelaplace transform of e 2t. Thus L n u 2 (t)e 2t o = e 2s 4 s +2 : c. Let G(s) = Lfg(t)g. Let Y (s) be the Laplace transform of the solution to the given IVP. Then we have (s 2 +5s +6)Y (s) = G(s) so that 1 Y (s) = G(s) s 2 +5s +6 ψ 1 = G(s) (s + 2)(s +)! : By a simple partial-fractions computation, we have 1 Y (s) = G(s) s +2 1 : s +

7 Let H(s) = 1 s +2 1 s + and h(t) = L 1 fh(s)g = e 2t e t : Then the solution y(t) is the convolution of h(t) with g(t). That is y(t) = = Z t 0 Z t 0 g(t fi)h(fi) dfi g(t fi)(e 2fi e fi ) dfi: 7. Consider the initial value problem y 0 =1+y 2 ; y(0) = 0. Let y = a 1 x + a 2 x 2 + a x + a 4 x 4 + be the solution. (Since y(0) = 0, we may leave the term a 0 out of the power series.) a. Recall that two convergent power series may be multiplied as though they were polynomials. Multiply the above expression for y by itself to find the first five non-zero terms of the power series for y 2. b. Now equate the power series for 1 + y 2 and y 0 to find the values of a 1 through a 6. Write down the beginning of the power series for y. (As you may have noticed, the solution to this IVP is y =tan(x), so you've just found the beginning of the power series for tangent.) Solution: a. We have y 2 = a 2 1 x2 + a 1 a 2 x + a 1 a x 4 + a 1 a 4 x 5 + a 1 a 5 x 6 + +a 2 a 1 x + a 2 2 x4 + a 2 a x 5 + a 2 a 4 x 6 + +a a 1 x 4 + a a 2 x 5 + a 2 x6 + +a 4 a 1 x 5 + a 4 a 2 x 6 + +a 5 a 1 x 6 + = a 2 1 x2 +2a 1 a 2 x +(2a 1 a + a 2 2)x 4 +(2a 1 a 4 +2a 2 a )x 5 +(2a 1 a 5 +2a 2 a 4 + a 2 )x 6 +

8 b. We have and 1+y 2 = 1+0x + a 2 1 x2 +2a 1 a 2 x +(2a 1 a + a 2 2)x 4 +(2a 1 a 4 +2a 2 a )x 5 +(2a 1 a 5 +2a 2 a 4 + a 2 )x 6 + y 0 = a 1 +2a 2 x +a x 2 +4a 4 x +5a 5 x 4 +6a 6 x 5 +7a 7 x 6 + : Equating coefficients, we get a 1 =1; 2a 2 =0,sothat a 2 =0; a = a 2 1 =1,so that a = 1 ; 4a 4 =2a 1 a 2 =0,sothat a 4 =0; 5a 5 =2a 1 a + a 2 2 = 2, so that a 5 = 2 15 ; 6a 6 =2a 1 a 4 +2a 2 a =0,sothat a 6 =0. The solution y begins y = x + 1 x x5 + :