MATH 286 Section D1 & X1: Assignment 2Solutions for the graded problems.Section 1.5, #14 [1 point] We are given the IVPy0+ cos(x)y = cos x, y(π ) = 2.Letting P(x) =Rcos(x)dx = sin x , the integrating factor becomeseP(x)= esin x.Multiplying by eP, we get(esin xy)0= esin xcos x =⇒ esin xy =Zesin xcos xdx + C =Zesin xcos xdx + C=⇒ y(x) = e− sin xesin x+ Ce− sin x+ e− sin xesin x= 1 + Ce− sin x.Since y(π) = 2, we get 1 + Ce0= 2 =⇒ C = 1.Section 1.5, #28 [2 points] We are given the ODE(1 + 2xy)dydx= 1 + y2.Viewing y as the independent variable and x as the dependent variable, the chain rulegivesdxdy=dydx−1,so the ODE rearranges todxdy=1 + 2yx1 + y2⇐⇒dxdy−2y1 + y2x =11 + y2,which is linear. The integrating factor is eR−2y1+y2dy= e− ln(1+y2)=11+y2. Multiplying by11+y2, we get11 + y2x0=1(1 + y2)2=⇒11 + y2x =Z1(1 + y2)2dy + C =12y1 + y2+ arctan y + C=⇒ x =12y + (1 + y2) arctan y + C(1 + y2).1Section 1.6, #4 [2 points] Making the substitution v = y/x, we get(x − vx)(xv)0= x + xv =⇒ v + xv0=1 + v1 − v=⇒ xdvdx=1 + v21 − v=⇒1 − v1 + v2dv =dxx=⇒ arctan v −12ln(1 + v2) = ln |x| + C=⇒ arctanyx−12ln(1 +y2x2) = ln |x| + C.Section 1.6, #38 [2 points] The equation(x + arctan y)| {z }Mdx +x + y1 + y2| {z }Ndy = 0is exact becauseMy=11 + y2= Nx.We solve this exact equation by finding an implicit solution of the form F(x, y) = 0. Tofind F, F must satisfyFx= M = x + arctan y, Fy= N =x + y1 + y2.This givesF =ZMdx = x2/2 + x arctan y + g(y),andFy= x/(1 + y2) + g0(y) = N =x + y1 + y2.Thus g0(y) =y1+y2, givingF(x, y) = x2/2 + x arctan y +12ln(1 + y2) + C.Section 1.6, #50 [2 points] The equationy00= (x + y0)2is second order, so we will try to reduce it to an equivalent first order equation. Since ydoes not appear in the equation, we substitute z = y0, and getz0= (x + z)2.2This new ODE is first order. To solve it, we make a linear substitution v = x + z to getv0= 1 + z0= 1 + v2.This last equation is separable, and we getdvv2+ 1= dx =⇒ arctan(v) = x + C1=⇒ v = tan(x + C1) =⇒ z = tan(x + C1) − x.We now integrate z = y0to get y:y(x) =Z(tan(x + C1) − x)dx + C2= ln | sec(x + C1)| −x22+ C2.The above integral was computed using the tables in the course textbook. Note that thereare two constants of integration here because the original equation was second order.Additional Problem [1 point] Let˜P(x) be any other antiderivative of p(x). I.e., P0(x) =p(x) =˜P0(x). Then˜P and P only differ by some additive constant of integration: Say˜P(x) = P(x) + A . Now, if˜y(x) denotes the general solution to the ODE obtained byusing˜P instead of P, we get (for some arbitrary constant K)˜y(x) = e−˜P(x)Ze˜P(x)q(x)dx + K= eP(x)−AZeP(x)+Aq(x)dx + K= eP(x)ZeP(x)q(x)dx +KeA.This last expression is exactly the one for y(x) once we take C
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