Solutions to Homework 3

Transcription

1 Solutions to Homework 3 Section 3.4, Repeated Roots; Reduction of Order Q 1). Find the general solution to 2y + y = 0. Answer: The charactertic equation : r 2 2r + 1 = 0, solving it we get r = 1 as a repeated root, so the general solution given by y(t) = c 1 e t + c 2 te t. Q 2). Find the general solution to 9 + 6y + y = 0 Answer: The charactertic equation : 9r 2 + 6r + 1 = 0, solving it we get r = 1/3 as a repeated root, so the general solution given by y(t) = c 1 e t/3 + c 2 te t/3. Q 8). Find the general solution to y + 9y = 0 Answer: The charactertic equation : 16r r + 9 = 0, solving it we get r = 3/4 as a repeated root, so the general solution given by y(t) = c 1 e 3t/4 + c 2 te 3t/4 Q 11). Solve for the given initial value problem 9 12y + 4y = 0 with y(0) = 2 and y (0) = 1, and sketch the graph of the solution and describe its behavior of increasing t. Answer: The charactertic equation : 9r 2 12r + 4 = 0, 1

2 solving it we get r = 2/3 as a repeated root, so the general solution given by So Putting in the initial condition we get y(t) = c 1 e 2t/3 + c 2 te 2t/3. y (t) = (2c 1 /3 + c 2 + 2tc 2 /3)e 2t/3. y(0) = c 1 = 2, y (0) = 2c 1 /3 + c 2 = 1. Solving it we get c 1 = 2 and c 2 = 7/3, hence the solution y(t) = 2e 2t/3 7 3 te2t/3. To see the graph of the solution, one can type the following commands in MATLAB: syms t; ezplot( 2 exp(2 t/3) 7/3 t exp(2 t/3) ); The solution y(t) when t. Q 12). Solve for the given initial value problem 6y + 9y = 0 with y(0) = 0 and y (0) = 2, and sketch the graph of the solution and describe its behavior of increasing t. Answer: The charactertic equation : r 2 6r + 9 = 0, solving it we get r = 3 as a repeated root, so the general solution given by So Putting in the initial condition we get y(t) = c 1 e 3t + c 2 te 3t. y (t) = (3c 1 + c 2 + 3tc 2 )e 3t. y(0) = c 1 = 0, y (0) = 3c 1 + c 2 = 2. Solving it we get c 1 = 0 and c 2 = 2, hence the solution y(t) = 2te 3t. To see the graph of the solution, one can type the following commands in MATLAB: syms t; ezplot( 2 t exp(3 t) ); The solution y(t) when t. Q 14). Solve for the given initial value problem + 4y + 4y = 0 with y( 1) = 2 and y ( 1) = 1, and sketch the graph of the solution and describe its behavior of increasing t. 2

3 Answer: The charactertic equation : r 2 + 4r + 4 = 0, solving it we get r = 2 as a repeated root, so the general solution given by So Putting in the initial condition we get y(t) = c 1 e 2t + c 2 te 2t. y (t) = ( 2c 1 + c 2 2tc 2 )e 2t. y(0) = c 1 e 2 c 2 e 2 = 2, y ( 1) = ( 2c 1 + c 2 + 2c 2 )e 2 = 1. Solving it we get c 1 = 7e 2 and c 2 = 5e 2, hence the solution y(t) = 7e 2 2t + 5te 2 2t. To see the graph of the solution, one can type the following commands in MATLAB: syms t; ezplot( 7 exp( 2 2 t) + 5 t exp( 2 2 t) ); The solution y(t) first increase, decrease, and y(t) when t. Q 23). Use the method of reduction to find a second solution: t 2 4ty + 6y = 0, y > 0; y 1 (t) = t 2. Answer: Assume y 2 (t) = v(t)y 1 (t), and y 2(t) = v (t)y 1 (t) + v(t)y 1(t) = v (t)t 2 + 2tv(t), 2 (t) = v (t)t 2 + 2tv (t) + 2v(t) + 2tv (t) = v (t)t 2 + 4tv (t) + 2v(t). Insert y 2, 2 and y 2 back into the ODE, we get that Thus we have t 2 (v (t)t 2 + 4tv (t) + 2v(t)) 4t(v (t)t 2 + 2tv(t)) + 6v(t)t 2 = 0, t 4 v (t) = 0. v (t) = 0. Let u = v, we have u (t) = 0, so u(t) = C 1. Thus, solving it, we get v = u = C 1, v(t) = C 1 t + C 2. Now take C 1 = 1 and C 2 = 0, we get v(t) = t, and y 2 (t) = v(t)y 1 (t) = t 3 3

4 Q 25). Use the method of reduction to find a second solution: t 2 + 3ty + y = 0, y > 0; y 1 (t) = t 1. Answer: Assume y 2 (t) = v(t)y 1 (t), and y 2(t) = v (t)y 1 (t) + v(t)y 1(t) = v (t)t 1 v(t)t 2, 2 (t) = v (t)t 1 v (t)t 2 v (t)t 2 + 2v(t)t 3 = v (t)t 1 2v (t)t 2 + 2v(t)t 3. Insert y 2, 2 and y 2 back into the ODE, we get that t 2 (v (t)t 1 2v (t)t 2 + 2v(t)t 3 ) + 3t(v (t)t 1 v(t)t 2 ) + v(t)t 1 = 0, tv (t) + v (t) = 0. Let u = v, we have tu (t) + u(t) = 0, By separation of variables, u(t) = C 1 /t. Thus, solving it, we get v = u = C 1 /t, v(t) = C 1 ln(t) + C 2. Now take C 1 = 1 and C 2 = 0, we get v(t) = ln(t), and y 2 (t) = v(t)y 1 (t) = t 1 ln(t) Section 3.5, Nonhomogeneous Equations; Method of Undetermined Coefficients Q 1). Find the general solution to 2y 3y = 3e 2t. Answer: The charactertic equation to the homogeneous ODE 2y 3y = 0 r 2 2r 3 = 0, with roots r 1 = 3 and r 2 = 1. So the general solution to y h (t) = c 1 e 3t + c 2 e t. 2y 3y = 3e 2t y p (t) = Ae 2t, 4

5 y p(t) = 2Ae 2t, p = 4Ae 2t. Hence, putting y p, y p and p back into we get 4Ae 2t 2 2Ae 2t 3Ae 2t = 3e 2t. Th gives us an algebraic equation 3A = 3. Solving for A gives us A = 1. So the general solution to (N) y(t) = c 1 e 3t + c 2 e t e 2t. Q 6). Find the general solution to + 2y = sin(2t). Answer: The charactertic equation to the homogeneous ODE + 2y = 0 r 2 + 2r = 0, with roots r 1 = 2 and r 2 = 0. So the general solution to y h (t) = c 1 e 2t + c y = sin(2t) y p (t) = At + B 1 sin(2t) + B 2 cos(2t), y p(t) = A + 2B 1 cos(2t) 2B 2 sin(2t), p = 4B 1 sin(2t) 4B 2 cos(2t). Hence, putting y p, y p and p back into we get 4B 1 sin(2t) 4B 2 cos(2t) + 2(A + 2B 1 cos(2t) 2B 2 sin(2t)) = sin(2t). Th gives us an three algebraic equations 2A = 3, Solving them gives us 4B 1 4B 2 = 4, 4B 2 + 4B 1 = 0. A = 3 B 1 = B 2 =

6 So the general solution to (N) y(t) = c 1 e 2t + c 2 + 3t sin(2t) 1 2 cos(2t). Q 8). Find the general solution to + 2y + y = 2e t. Answer: The charactertic equation to the homogeneous ODE + 2y + y = 0 r 2 + 2r + 1 = 0, with repeated roots r = 1. So the general solution to y h (t) = c 1 e t + c 2 te t. + 2y + y = 2e t y p (t) = At 2 e t, y p(t) = (2At At 2 )e t, p = (2A 2At 2At + At 2 )e t. Hence, putting y p, y p and p back into we get Simplify it, we get (2A 2At 2At + At 2 )e t + 2(2At At 2 )e t + At 2 e t = 2e t. 2Ae t = 2e t. Th gives us an three algebraic equations 2A = 2. Solving them gives us A = 1. So the general solution to (N) y(t) = (c 1 + c 2 t + t 2 )e t. Q 9). Find the general solution to 2 + 3y + y = t sin t. Answer: The charactertic equation to the homogeneous ODE 2 + 3y + y = 0 2r 2 + 3r + 1 = 0, 6

7 with roots r 1 = 1 and r 2 = 1/2. So the general solution to y h (t) = c 1 e t + c 2 e t/ y + y = t sin t y p (t) = A 1 t 2 + A 2 t + A 3 + B 1 sin t + B 2 cos t, y p(t) = 2A 1 t + A 2 + B 1 cos t B 2 sin t, p = 2A 1 B 1 sin t B 2 cos t. Hence, putting y p, y p and p back into we get Th gives us algebraic equations Solving them gives us So the general solution to (N) 2(2A 1 B 1 sin t B 2 cos t) + 3(2A 1 t + A 2 + B 1 cos t B 2 sin t) +A 1 t 2 + A 2 t + A 3 + B 1 sin t + B 2 cos t = t sin t. A 1 = 1, 6A 1 + A 2 = 0, 4A 1 + 3A 2 + A 3 = 0, 2B 1 3B 2 + B 1 = 3, 2B 2 + 3B 1 + B 2 = 0. A 1 = 1, A 2 = 6, A 3 = 14, B 1 = 3 10, B 2 = y(t) = c 1 e t + c 2 e t/2 + t 2 6t sin t 9 cos t. 10 Q 16). Solve the initial value problem + 4y = t 2 + 3e t, with y(0) = 0 and y (0) = 2. Answer: The charactertic equation to the homogeneous ODE + 4y = 0 r = 0, 7

8 with complex roots r = ±2i. So the general solution to y h (t) = c 1 sin(2t) + c 2 cos(2t). + 4y = t 2 + 3e t y p (t) = A 1 t 2 + A 2 t + A 3 + Be t, y p(t) = 2A 1 t + A 2 + Be t, y p = 2A 1 + Be t. Hence, putting y p, y p and y p back into we get Th gives us algebraic equations Solving them gives us So the general solution to (N) Hence 2A 1 + Be t + 4(A 1 t 2 + A 2 t + A 3 + Be t ) = t 2 + 3e t 4A 1 = 1, 4A 2 = 0, 2A 1 + 4A 3 = 0, B + 4B = 3. A 1 = 1 4, A 2 = 0, A 3 = 1 8, B 2 = 3 5. y(t) = c 1 sin(2t) + c 2 cos(2t) + t et 5. y (t) = 2c 1 cos(2t) 2c 2 sin( 2t) + t 2 + 3et 5. With the initial conditions y(0) = 0 and y (0) = 2, we have y(0) = c = 0, y (0) = 2c = 2. Solving for c 1 and c 2, we get c 1 = 7 10, c 2 = 19 40, so the solution y(t) = 7 19 sin(2t) cos(2t) + t et 5. 8

9 Q 17). Solve the initial value problem 2y + y = te t + 4, with y(0) = 1 and y (0) = 1. Answer: The charactertic equation to the homogeneous ODE 2y + y = 0 r 2 + 2r + 1 = 0, with repeated roots r = 1. So the general solution to y h (t) = c 1 e t + c 2 te t. 2y + y = te t + 4 y p (t) = At 3 e t + B, y p(t) = (3At 2 + At 3 )e t, y p = (3At 2 + At 3 + 3At 2 + 6At)e t. Hence, putting y p, y p and y p back into we get (3At 2 + At 3 + 3At 2 + 6At)e t 2(3At 2 + At 3 )e t + At 3 e t + B = te t + 4. Th gives us algebraic equations Solving them gives us 6A = 1, B = 4. A = 1 6, B = 4. So the general solution to (N) y(t) = (c 1 + c 2 t t3 )e t + 4. We have y (t) = (c 1 + c 2 t t3 + c t2 )e t. So the initial conditions y(0) = 1 and y (0) = 1 gives us c = 1, c 1 + c 2 = 1. Solving for it we get So the solution to the I.V.P c 1 = 3, c 2 = 4. y(t) = ( 3 + 4t t3 )e t