# M427K (54315) Midterm #1 Solutions

## Transcription

M427K (54315) Midterm #1 Solutions

M427K (54315) Midterm #1 Solutions 1. Find the general solution of the following first order ODEs. −2t (a) t dy dt + 2ty = e Solution. Write the linear equation in standard form: dy 1 + 2y = e−2t . dt t Identify p(t) = 2 and compute the integrating factor R µ(t) = e 2 dt = e2t . Multiply by the integrating factor to obtain d 2t 1 ye = . dt t Integrating: ye2t = ln |t| + C and the general solution is y = e−2t (ln |t| + C). (b) dy dt = e−2y sin(3t) Solution. The equation is separable, and can be expressed as Z Z 2y e dy = sin(3t) dt. Integrating, 1 2y 1 e = − cos(3t) + C. 2 3 Solving for y: y= 1 2 ln − cos(3t) + 2C . 2 3 2. Consider the ODE dy = y 2 − 1. dt (a) Sketch the right-hand side f (y) = y 2 − 1. Solution. (b) Draw the phase line. Solution. 1 (c) Find all equilibrium points and classify each one as stable, unstable, or semistable. Solution. There is an unstable equilibrium at y = 1 and a stable equilibrium at y = −1. (d) Sketch several solutions of the ODE that exhibit the different types of behaviour that can be expected. Solution. 3. Consider the ODE x2 y 00 − 5xy 0 + 9y = 0. (a) Verify that y1 = x3 is a solution of this ODE. Solution. Compute the derivatives of y1 : y10 = 3x2 , y100 = 6x. Substitute these into the ODE: x2 y100 − 5xy10 + 9y1 = x2 (6x) − 5x(3x2 ) + 9(x3 ) = 0. Thus y1 is a solution. (b) Use the method of reduction of order to find a second solution y2 of the ODE. Solution. Look for a solution of the form y2 = x3 v. The derivatives are y20 = x3 v 0 + 3x2 v, y200 = x3 v 00 + 6x2 v 0 + 6xv. Substitute into the ODE: x2 y200 − 5xy20 + 9y2 = x2 (x3 v 00 + 6x2 v 0 + 6xv) − 5x(x3 v 0 + 3x2 v) + 9x3 v = x5 v 00 + x4 v 0 = 0. Substitute w = v 0 : x5 w0 + x4 w = 0. This is separable so we can write Z Z dx dw =− . w x Integrate this to obtain ln w = − ln x, Then we can obtain Z v= ⇒ w= 1 . x 1 dx = ln |x| . x Thus a second solution is y2 = y1 v = x3 ln |x| . 4. Solve the following initial value problem using any of the techniques learned in this class. y 00 − 2y 0 + 2y = sin t, y(0) = y 0 (0) = 1. Solution. The characteristic equation is r2 − 2r + 2 = 0 which has the roots r= 2± √ 4−8 = 1 ± i. 2 The homogeneous solution is yh (t) = C1 et cos t + C2 et sin t. Look for a particular solution of the form yp (t) = A cos t + B sin t, yp0 (t) = −A sin t + B cos t, yp00 (t) = −A cos t − B sin t. Substitute this into the ODE: yp00 − 2yp0 + 2yp = (A − 2B) cos t + (2A + B) sin t. Require this to equal sin t, which yields A − 2B = 0 and 2A + B = 1. These have the solution A = 2/5, B = 1/5. The general solution is 2 1 y(t) = yh (t) + yp (t) = C1 et cos t + C2 et sin t + cos t + sin t. 5 5 The derivative of this is y 0 (t) = C1 et (cos t − sin t) + C2 et (sin t + cos t) − 2 1 sin t + cos t. 5 5 The IC require 2 1 = 1, y 0 (0) = C1 + C2 + = 1. 5 5 The solutions are C1 = 3/5 and C2 = 1/5. Thus the solution of the IVP is 1 2 1 3 y(t) = et cos t + et sin t + cos t + sin t. 5 5 5 5 y(0) = C1 +