# M427K (55590) Midterm #1 Solutions

## Transcription

M427K (55590) Midterm #1 Solutions

M427K (55590) Midterm #1 Solutions 1. Consider the linear ODE t dy + y = t2 . dt (a) Suppose you are given the initial condition y(t0 ) = y0 . Find all values of t0 that guarantee that the initial value problem has a unique solution. Solution. First put the ODE into standard form: dy 1 + y = t. dt t We can identify p(t) = 1t , which is continuous everywhere except at t = 0, and g(t) = t, which is continuous everywhere. Thus choosing t0 to be any non-zero value will guarantee that the IVP has a unique solution. (b) Suppose t0 = 2. Find the largest interval on which the initial value problem is guaranteed to have a unique solution. Solution. Since t0 = 2 > 0, the IVP will have a unique solution on the interval (0, ∞). (c) Find the general solution of the ODE. Solution. The equation needs to be in standard form, as in part (a). This is a linear first-order ODE, so we look for an integrating factor. Z µ(t) = exp 1 dt = eln t = t. t Multiply by µ(t): d (ty) = t2 . dt Integrate: ty = 1 3 t + C. 3 Solve for y: y(t) = 1 2 C t + . 3 t 2. Consider the ODE dy = −y 2 . dt (a) Sketch the right-hand side f (y) = −y 2 . Solutions. 1 (b) Draw the phase line. Solutions. (c) Find all equilibrium points and classify each one as stable, unstable, or semistable. The function f (y) vanishes at y = 0, so y = 0 is an equilibrium. From the phase plot, it is a semistable equilibrium. Solutions. (d) Sketch several solutions of the ODE that exhibit the different types of behaviour that can be expected. 2 3. Consider the ODE −t − y 2 + dy = 0. dt (a) Is the equation linear or nonlinear? Solution. The equation is nonlinear since it contains the term y 2 . (b) Is the equation separable? Solution. The equation is not separable since it involves a sum of a function of t and a function of y. (c) Is the equation exact? Justify your answer. Solution. To check if the equation is exact, identify M (t, y) = −t − y 2 and N (t, y) = 1. Then compute My = −2y, Nt = 0. These are not equal so the equation is not exact. (d) Given the initial condition y(0) = 1, use Euler’s method with h = 0.1 to approximate y(0.1). Solution. The equation can be rewritten as dy = t + y2 dt so that f (t, y) = t + y 2 . We also know that t0 = 0 and y0 = 1. Euler’s method involves computing yn+1 = yn + h(tn + yn2 ). Thus y(0.1) ≈ y1 = 1 + 0.1(0 + 12 ) = 1.1. 4. Solve the following initial value problem using any of the techniques learned in this course. y 00 − 3y 0 + 2y = 6e−t , y(0) = 1, y 0 (0) = 2 Solution. The characteristic equation for this problem is r2 − 3r + 2 = (r − 1)(r − 2) = 0, which has the two real roots r = 1, 2. Thus the homogeneous solution is yh (t) = C1 et + C2 e2t . Now we look for a particular solution of the form yp (t) = Ae−t , yp0 (t) = −Ae−t , 3 yp00 (t) = Ae−t . Substitute this into the ODE: yp00 (t) − 3yp0 (t) + 2yp (t) = (A + 3A + 2A)e−t = 6Ae−t . Setting this equal to the given right-hand side 6e−t , we obtain A = 1. Thus the general solution is y(t) = yh (t) + yp (t) = C1 et + C2 e2t + e−t . We we also need the derivative of this: y 0 (t) = C1 et + 2C2 e2t − e−t . The initial conditions require y(0) = C1 + C2 + 1 = 1 ⇒ C2 = −C1 , 0 y (0) = C1 + 2C2 − 1 = 2 ⇒ −C1 = 3, which yields C1 = −3 and C2 = 3. Thus the solution of the IVP is y(t) = −3et + 3e2t + e−t . 4