# MATH 391 Test 1 Fall, (1) (12 points each)compute the general solution of each of the following differential equations: = 4x 2y.

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1 MATH 391 Test 1 Fall, 2018 (1) (12 points each)compute the general solution of each of the following differential equations: (a) (b) x dy dx + xy = x2 + y. (x + y) dy dx = 4x 2y. (c) yy + (y ) 2 = 0 (y is a function of x). dy (a) Linear: dx + [1 1 x ]y = x. µ = exp[ (1 1 )dx] = x 1 e x. So [x 1 e x y] = e x. The solution is x 1 e x y = e x + C. So y = x + Cxe x. (b) Homogeneous: Use z = y/x. x dz 1+z (z+4)(z 1) = A z+4 + B z 1. dx = 4 2z 1+z z = 4 3z z2 1+z = (z+4)(z 1) 1+z. So 1+z = A(z 1)+B(z+4). Substitute z = 1, 4 to get A = 3/5, B = 2/5. (3/5) ln((y/x)+4)+(2/5) ln((y/x) 1) = ln(x)+c. You need not simplify, but ln[(y + 4x) 3 x 2 ] + ln[(y x) 2 x 3 ] = C. (c) Reduction of Order: y = v, y = v dv dy dv and so yv and so y = C are solutions, but you need not include them]. dv v ln v = ln y + C 1 or dy dx = v = C 1y 1. So y 2 /2 = C 1 x + C 2. dy + v2 = 0. [v = 0 = dy y and so (a) (b) (2) (12 points each) Solve the following initial value problems: x 2 y dy dx + xy2 = x + 1. with y(3) = 1. dy dx = xy 2 + x 2 y 2. with y(0) = 1. (a) (x 2 y)dy + (xy 2 x 1)dx = 0 is exact with F = 1 2 x2 y 2 + H(x) and so xy 2 + H (x) = xy 2 x 1. So F = 1 2 [x2 y 2 x 2 2x]. The general solution is F = C. So 1 2 [ ] = C and 3 = C. Solution x 2 y 2 x 2 2x = 6. (b) Variables separable with y 2 dy = [x+x 2 ]dx. So y 1 = x 2 /2+x 3 /3+C. Since y = 1 when x = 0, C = 1.

2 (3) (10 points each)in the following equations y is a function of t. (a) Solve the initial value problem 2y + 3y 5y = 0, y(0) = 14, y (0) = 0. (b) Compute the general solution of 2y + 3y + 5y = 0. (a) The characteristic equation is 0 = 2r 2 + 3r 5 = (2r + 5)(r 1) with roots r = 5/2, 1. The general solution is y = C 1 e 5t/2 + C 2 e t. So 14 = y(0) = C 1 + C 2 and 0 = y (0) = 5C 1 /2 + C 2. Subtracting, 14 = 7C 1 /2 and so C 1 = 4, C 2 = 10. y = 4e 5t/2 + 10e t. (b) The characteristic equation is 0 = 2r 2 + 3r + 5 with roots the complex conjugate pair r = ( 3/4) ± i 31/4. The general solution is y = C 1 e 3t/4 cos( 31t/4) + C 2 e 3t/4 cos( 31t/4). (4)(10 points) Assume that y 1, y 2 are solutions of the equation y +py +qy = 0 where p and q are functions of t defined for all t. (a) Prove Abel s Theorem, that the Wronskian of of the pair y 1, y 2 satisfies a first order equation. (b) Explain why the solution of the equation in (a) shows that either the Wronskian is identically zero or never zero. q y 1 y 2 y 2 y 1 = 0 p y 1 y 2 y 2 y 1 = W 1 y 1 y 2 y 2 y 1 = W Adding we obtain 0 = pw + W with solution W = Cexp[ p(t)dt]. The exponential is positive and so if C 0 the Wronskian is never zero. (5) (10 points) I borrow \$10, 000 which has an interest rate of 7% per year, compounded continuously. I pay it off continuously at a rate of \$100 per month. Set up an initial value problem (differential equation and initial conditions) whose solution is the quantity P (t) of dollars that I have left to pay time t (until the loan is paid off). You need not solve the equation. The payoff rate is \$1200 per year. dp dt =.07P 1200 with P (0) = 10, 000.

3 MATH 391 Test 2 Fall, 2018 (1)(18 points) Solve the initial value problem: y 3y = 18t + 36e 3t with y(0) = 0 and y (0) = 0. Characteristic equation for the homogeneous is r 2 3r = 0 with roots r = 0, 3. So y h = C 1 + C 2 e 3t. 18t has associated root 0 and 36e 3t has associated root 3. Yp 1 = (At + B) + (Ce 3t ) but we must multiply the first block by t. 0 [Y p = At 2 + Bt + + Ce 3t ], 3 [Y p = 2At + B 3Ce 3t ], 1 [Y p = + + 2A + 9Ce 3t ]. 18t + 36e 3t = 6At + ( 3B + 2A) + 18Ce 3t. So A = 3, B = 2, C = 2. y g = C 1 + C 2 e 3t 3t 2 2t + 2e 3t, y g = +3C 2 e 3t 6t 2 6e 3t. 0 = C 1 + C 2 + 2, 0 = 3C So C 2 = (8/3), C 1 = (14/3). y = (14/3) + (8/3)e 3t 3t 2 2t + 2e 3t. (2)(18 points) Consider the sixth order differential equation y (6) +2y +y = 3t 2 + te t/2 7e t/2 sin(( 3/2)t) + cos(( 3/2)t) + 1. (a) Compute the general solution of the associated homogeneous equation. (b) Write down the test function with the fewest terms which can be used to obtain a particular solution via the Method of Undetermined Coefficients. Do not solve for the constants. Characteristic equation for the homogeneous is r 6 + 2r = (r 3 + 1) 2 = 0. Since r 3 +1 = (r +1)(r 2 r +1) the roots are 1, (1/2)±( 3/2)i each repeated twice. y h = C 1 e t + C 2 te t + C 3 e (1/2)t cos(( 3/2)t) + +C 4 e (1/2)t sin(( 3/2)t)+ C 5 te (1/2)t cos(( 3/2)t) + C 6 te (1/2)t sin(( 3/2)t). Associated root for 3t 2 and 1 is 0, for te t/2 is (1/2), for 7e t/2 sin(( 3/2)t) is (1/2) ± ( 3/2)i and for cos(( 3/2)t) is ±( 3/2)i. Y 1 p = (At 2 + Bt + C) + (Dt + E)e t/2 + (F e t/2 cos(( 3/2)t) + Ge t/2 sin(( 3/2)t)) + (H cos(( 3/2)t) + I sin(( 3/2)t)).

4 Since (1/2) ± ( 3/2)i is repeated twice we must multiply the third block by t 2. Y p = (At 2 + Bt + C) + (Dt + E)e t/2 + t 2 (F e t/2 cos(( 3/2)t)+Ge t/2 sin(( 3/2)t))+(H cos(( 3/2)t)+I sin(( 3/2)t)). (3)(16 points) Compute the general solution of y + y = tan(t). Characteristic equation for the homogeneous is r = 0 with roots ±i and so y h = C 1 cos(t) + C 2 sin(t). We look for y p = u 1 cos(t) + u 2 sin(t). u 1(cos(t)) + u 2(sin(t)) = 0 u 1( sin(t)) + u 2(cos(t)) = tan(t). Wronskian is 1. u 1 = sin(t) tan(t) = sin 2 (t)/ cos(t) = (cos 2 (t) 1)/ cos(t) = cos(t) sec(t), u 2 = cos(t) tan(t) = sin(t). u 1 = sin(t) ln sec(t) + tan(t), u 2 = cos(t), y g = C 1 cos(t) + C 2 sin(t) sin(t) ln sec(t) + tan(t). (4)(16 points) For the differential equation, (x 1)y xy +y = 0, y 1 = e x is a solution. Use the method of Reduction of Order to compute a second solution y 2 which is independent of the first one. y 2 (x) = ue x. 1 [y 2 = ue x ] x [y 2 = ue x + u e x ] (x 1) [y 2 = ue x + 2u e x + u e x ] 0 = 0 + u (x 2)e x + u (x 1)e x (x 1)v = (x 2)v, and so dv v = x 1 dx. u = v = (x 1)e x, and so u = xe x and y 2 = uy 1 = x. (5)(14 points) A hanging spring is stretched 6 inches (=.5 feet) by a weight of 24 pounds. (a) Set up the initial value problem (differential equation and initial conditions) which describes the motion, neglecting friction, when the weight is pulled down an additional foot and is then released and is subjected to an external force of 6 cos(ωt). You need not solve the equation. (Recall that g, the acceleration due to gravity is 32 feet/second 2.)

5 (b) Write down the test function with the fewest terms which can be used to obtain a particular solution via the Method of Undetermined Coefficients when ω = 5. (c) Write down the test function with the fewest terms which can be used to obtain a particular solution via the Method of Undetermined Coefficients when ω is the resonance frequency. (a) w = 24 = mg = 32m, so m = 3/4. 24 = w = k L = k(1/2), so k = 48. c = 0. (3/4)y + 48y = 6 cos(ωt), or, y + 64y = 8 cos(ωt), y(0) = 1, y (0) = 0. (b) y p = A cos 5t + B sin 5t. (c) The natural frequency is ω = 8. y p = At cos 8t + Bt sin 8t. (6) (18 points) For the differential equation: (2 x 2 )y + x 2 y + (x 2x 3 )y = 0 Compute the recursion formula for the coefficients of the power series solution centered at x 0 = 0. Use it to compute the first three nonzero terms of the series for the solution with y(0) = 24 and y (0) = 0. 2y = Σ 2n(n 1)a n x n 2 [k = n 2] = Σ2(k + 2)(k + 1)a k+2 x k. x 2 y = Σ n(n 1)a n x n [k = n] = Σ k(k 1)a k x k. x 2 y = Σ na n x n+1 [k = n + 1] = Σ (k 1)a k 1 x k. xy = Σ a n x n+1 [k = n + 1] = Σ a k 1 x k. 2x 3 y = Σ 2a n x n+3 [k = n + 3] = Σ 2a k 3 x k. a k+2 = 1 2(k+2)(k+1) [k(k 1)a k (k 1)a k 1 a k 1 +2a k 3 ]. a 0 = 24, a 1 = 0. Recursion formula: a k+2 = 1 2(k + 2)(k + 1) [k(k 1)a k ka k 1 + 2a k 3 ]. k = 0 : a 2 = 1 4 [ ] = 0. k = 1 : a 3 = 1 12 [0 a 0 + 0] = 2. k = 2 : a 4 = 1 24 [2a 2 2a 1 + 0] = 0 k = 3 : a 5 = 1 40 [6a 3 3a 2 + 2a 0 ] = = y = x x

6 MATH 391 Test 3 Fall, 2018 (1) Compute the general solution for each of the differential equations, (y is a function of x in each case). (a) x 2 y + 7xy + 6y = 0, x > 0 (20 points) (b) y + 7y + 6y = 6x. (a) Euler Equation: Indicial Equation r(r 1) + 7r + 6 = r 2 + 6r + 6 = 0 with roots r = 3 ± 3. General solution is: y = C 1 x C 2 x 3 3. (b) For the homogeneous equation y + 7y + 6y = 0 the characteristic equation is r 2 + 7r + 6 = (r + 6)(r + 1) = 0 with roots r = 6, 1. So y h = C 1 e 6x + C 2 e x. The associated root for 6x is 0 and so Y p = Ax + B. Y p = A, Y p = 0. So Y p + 7Y p + 6Y p = 6Ax + (6B + 7A) = 6x. So A = 1, B = 7/6. y = C 1 e 6x + C 2 e x + x (7/6). (2) (a) State the definition of the Laplace transform and use it to compute to compute for a function f(t) = e t the Laplace Transform L(f)(s). (b) Compute the Laplace Transform L(y) for the solution of the initial value problem: (15 points) 3y 3y + 2y = e t, y(0) = 5 and y (0) = 2. (a) L(f)(s) = e st f(t) dt. With f(t) = e t, 0 L(f)(s) = (b) 0 e st e t dt = 0 e (s+1)t dt = 1 s + 1 e (s+1)t 0 = 1 s + 1. L(y ) = sl(y) y(0), L(y ) = sl(y ) y (0) = s 2 L(y) sy(0) y (0). L(3y 3y + 2y) = 3[s 2 L(y) s5 ( 2)] 3[sL(y) 5] + 2L(y) = 1 s + 1. So L(y) = [ 1 s s 21] [3s2 3s + 2]. (3) For the differential equation: xy (2 + x)y = 0

7 (a) The point x 0 = 2 is an ordinary point. Compute the recursion formula for the coefficients of the power series solution centered at x 0 = 2. (15 points) (b) Solve the equation explicitly, by using the method of Reduction of Order. (10 points) (c) The point x = 0 is a regular singular point. Compute the associated Euler equation and compute the recursion formula for the coefficients of the series solution centered at x 0 = 0 which is associated with the larger root.(20 points) (a) Let X = x+2 so that x = X 2. The equation becomes (X 2)y Xy = 0. y = Σa n X n. 2y = Σ 2n(n 1)a n X n 2 [k = n 2] = Σ 2(k + 2)(k + 1)a k+2 X k. +Xy = Σ n(n 1)a n X n 1 [k = n 1] = Σ (k + 1)ka k+1 X k. Xy = Σ na n X n [k = n] = Σ ka k X K Recursion formula: a k+2 = (k + 1)ka k+1 ka k. 2(k + 2)(k + 1) (b) Let v = y and v = y. x dv dx = (x + 2)v. dv ln(v) = v = x dx = x + 2 ln(x) + C 1. dy dx = C 1x 2 e x. So y = C 1 [x 2 e x 2xe x + 2e x ] + C 2. (c) Multiply by x to get x 2 y x(2+x)y = 0 with associated Euler equation x 2 y 2xy = 0. The indicial equation is r(r 1) 2r = r(r 3) = 0 with roots 0, 3. So that 3 is the larger root. y = x r Σa n x n = Σa n x n+r. x 2 y = Σ (n + r)(n + r 1)a n x n+r [k = n] = Σ(k + r)(k + r 1)a k x k+r. 2xy = Σ 2(n + r)a n x n+r [k = n] = Σ 2(k + r)a k x k+r. x 2 y = Σ (n + r)a n x n+r+1 [k = n + 1] = Σ (k 1 + r)a k 1 x k+r [(k + r)(k + r 1) 2(k + r)]a k = [(k + r)(k + r 3)]a k = (k 1 + r)a k 1. With r = 3 the Recursion Formula is: a k = (k + 2) k(k + 3) a k 1.

8 (4) (a) Compute the Fourier series for the function f(x) = x 2 on the interval [ 1, 1]. (b) Compute the solution u(x, t) for the partial differential equation on the interval [0, 1]: u t = 16u xx with u(0, t) = u(1, t) = 0 for t > 0 (boundary conditions) u(x, 0) = 3 sin(2πx) 5 sin(5πx) + sin(6πx). for 0 < x < 1 (initial conditions) (20 points) (a) y = x 2 is even and so b n = 0 for all n. a 0 = x2 dx = 2/3. a n = x 2 cos(nπx)dx = 2[(x 2 )( 1 nπ (b) L = 1, α = sin(nπx)) (2x)( cos(nπx))+(2)( (nπ) 2 (nπ) 3 sin(nπx))]1 0 = (0 0) + ( 4( 1)n 4( 1)n ((nπ) 2 0) + (0 0) = ) ((nπ) 2 ). y = (2/3) (1/2) + Σ 4( 1) n n=1 ((nπ) 2 ) cos(nπx). u(x, t) = 3e (8π)2t sin(2πx) 5e (20π)2t sin(5πx) + e (24π)2t sin(6πx).

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