MATH 286 Sections D1 & X1: Assignment 4Solutions for the graded problems.Section 3.5, #18 [2 points] For the ODELy = y(4)− 5y00+ 4y = ex− xe2x,the characteristic polynomial isP(r) = r4− 5r2+ 4 = (r2− 4)(r2− 1) = (r − 2)(r + 2)(r − 1)(r + 1).Since F(x) = ex− xe2x, our first guess for a particular solution might be˜yp(x) = Aex+ (Bx + C)e2x.But ex, e2xare both solutions to the homogeneous equation Ly = 0 (corresponding tothe multiplicity 1 roots r = 1, 2), so according to Rule 2, we need to multiply Aexand(Bx + C)e2xby x to eliminate any duplication between our particular solution and thecomplementary function yc. We thus take=⇒ yp(x) = Axex+ x(Bx + C)e2x.To find A, B, C, we use the fact that we must haveLyp= y(4)p− 5y00p+ 4yp= ex− xe2x.Comparing the coefficients of the xex, x2e2x, xe2xterms on both sides of this equation, wegetA =−24144, B =−6144, C =19144.Section 3.5, #26 [2 points] For the ODELy = y00− 6y0+ 13y = xe3xsin 2x,the characteristic polynomial isP(r) = r2− 6r + 13,which has roots r1,2= 3 ± 2i (by the quadratic formula). Therefore the complementaryfunction is yc(x) = c1e3xsin 2x + c2e3xcos 2x. Since F(x) = xe3xsin 2x, our first guess fora particular solution might be(Ax + B)e3xcos 2x + (Cx + D)e3xsin 2x.However, since the functions e3xsin 2x, e3xcos 2x are duplicated in the complementaryfunction yc, we must eliminate this duplication by multiplying the above expression byx. So the correct particular solution (according to Rule 2) becomesyp(x) = x(Ax + B)e3xcos 2x + x(Cx + D)e3xsin 2x.1Section 3.5, #28 [2 points] The characteristic polynomial for the ODELy = y(4)+ 9y00= (x2+ 1) sin 3xis P(r) = r4+ 9r2= r2(r + 3i)(r − 3i), which has roots ±3i and 0. Since the complexnumber a + ib = 3i (coming from the forcing term F(x) = (x2+ 1) sin 3x) is a rootof P(r), we have duplication between the complementary function and the part of thecandidate (Ax2+ Bx + C) cos 3x + (Dx2+ Ex + F) sin 3x. To eliminate this, we mustmultiply by x, to getyp(x) = (Ax3+ Bx2+ Cx) cos 3x + (Dx3+ Ex2+ Fx) sin 3x.Section 3.5, #54 [2 points] To apply Theorem 1 to the ODELy = y00+ y = csc2x,we must first find a pair of linearly independent solutions to the equation Ly = 0, whichwe easily find (by examining the characteristic polynomial) to be y1(x) = cos x andy2(x) = sin x. Then Theorem 1 tells us that a particular solution is given byyp= u1y1+ u2y2, u1(x) = −Zy2(x)F(x)W(x)dx, u2(x) =Zy1(x)F(x)W(x)dx.Here F(x) = csc2x, and W is the Wronskian W(x) = W(y1, y2)(x) = y1(x)y02(x) −y2(x)y01(x) = cos2(x) + sin2(x) = 1. Thus,u1(x) = −Zsin x csc x2dx = −Zcsc xdx = − ln |cscx − cot x|u2(x) =Zcos x csc x2dx =Zcot csc xdx = − csc x=⇒ yp(x) = − ln |cscx − cot x| cos x − 1,where we have evaluated the above antiderivatives using the integral tables in the text-book.Proof of Theorem 1 in Section 3.5 [2 points] Let y1, y2be a pair of linearly independentsolutions to the homogeneous equation Ly = 0 associated to the second order linearequation Ly = y00+ p(x)y0+ q(x)y = F(x), and let W(x) denote the Wronskian of y1andy2. Let u1and u2be the functions defined by the antiderivatives in the solution to theproblem above. Our goal is to show that yp= u1y1+ u2y2is a particular solution to theequation Ly = F.2To do this, we compute derivatives:y0p= u01y1+ u02y2+ u1y01+ u2y02=−Fy2y1W+Fy1y2W+ u1y01+ u2y02= u2y01+ u2y02.=⇒ y00p= u01y01+ u02y02+ u1y001+ u2y002=−Fy2y01W+Fy1y02W| {z }=FWW=F+u1y001+ u2y002= F + u1y001+ u2y002.ThereforeLyp= y00p+ p(x)y0p+ q(x)yp= F + u1y001+ u2y002+ u1p(x)y01+ u2p(x)y02+ u1q(x)y1+ u2q(x)y2= F + u1y001+ p(x)y01+ q(x)y1+ u2y002+ p(x)y02+ q(x)y2= F + u1Ly1+ u2Ly2= F.Thus ypis a particular solution.