MATH 286 Section D1 & X1: Assignment 1Solutions for the graded problems.Section 1.1, #16 [2 points] Subbing y(x) = erxinto the ODE 3y00+ 3y0−4y = 0 gives3r2erx+ 3rerx−4erx= 0 ⇐⇒ (3r2+ 3r −4)erx= 0.Since erx6= 0, we must have3r2+ 3r −4 = 0 ⇐⇒ r =−3 ±p9 −4 ·3 · (−4)2(3)= −12±√576.Section 1.2, #10 [1 point] Recall the integration by parts formula:Rudv = uv −Rvdu.Integrating by parts the ODEdydx= xe−x( with u = x and dv = e−xdx) givesy(x) =Zxe−xdx + C = −xe−x−Z(−e−x)dx + C = −xe−x− e−x+ C.The initial condition y(0) = 1 allows us to solve for C:y(0) = 0 −1 + C = 1 =⇒ C = 2.So the solution to the IVP isy(x) = −xe−x− e−x+ 2Section 1.2, #18 [2 points] We have x00(t) = a(t) = 50 sin 5t. Thereforex0(t) = −10 cos 5t + C1= v(t) and x(t) = −2 sin 5t + C1t + C2.Subbing in the ICs yields−10 = v(0) = −10 + C1=⇒ C1= 0 & 0 = x(0) = C2.So, x(t) = −2 sin 5t.Section 1.3, #14 [1 point] For the IVPy = y1/3, y(0) = 0,Theorem 1 does not guarantee the existence of a unique solution, since f (x, y) = y1/3iscontinous but not differentiable at the point (x0, y0) = (0, 0) corresponding to the giveninitial condition. (I.e., the hypotheses of the theorem are not satisfied by this example).Note that this IVP has more than one solution: y1(x) = 0 and y2(x) =2x33/2(whichcomes from separation of variables).1Section 1.4, #14 [2 points]dydx=1 +√x1 +√y⇐⇒ (1 +√y)dy = (1 +√x) dx.Integrating, we gety +23y3/2= x +23x3/2+ C.Section 1.4, #28 [2 points] The given ODE is separable, so by separating variables andintegrating:2√xdydx= cos2y ⇐⇒ sec2ydy =x−1/22dx=⇒ tan y = x1/2+ C.Using the IC y(4) = π/4, we calculate C:tan(π/4) = 1 = 41/2+ C =⇒ C = −1.Solving for y, we gety(x) = arctan(x1/2−1).Note: In the above example, we use the convention that√x means the positive squareroot of
View Full Document
Unlocking...