MT171: Question Sheet 11 - 2018/19: Hints to solutions Please see detailed solutions for question 1 in the document "Question 1 (QSheet 11)-detailed solution", posted in WEEK 11 of our MT171 Moodle page. yc(x) = C1e-T + c2e-2x for arb. consts. C1 and C2 . yp(x) = Asinx + B cosx for some consts. A and B , 1.(a) = (sin x - 3 cos x) , giving y(x) = C1e + cze-2ª + (sinx - 3 cosx) . (b) yc(x)= C1eª + C2e-3x for arb. consts. C1 and C2 . yp(x) = Axe" for some const. A = x giving y(x) = C1eª + C2e-3ª + xe". (c) 2 cosh x = eª + e-ª yc(x)= C1e" + C2e2ª + C3xe2ª for arb. consts. C1 , C2 and c3 . so taking yp(x) = Axe-" + Be" some consts. A and B = = + 2 , 3ex giving y(x)=C1e-T+ (c2 + C3x)e2ª + "e" + 3 3ex 2 . (d) yc(x) = ex(a1er + Q2e-ix) for arb. consts. Q1 and a2 , = e" (C1 cos x + C2 sin x) for arb. consts. C1 and C2 . yp(x) = xe" (Asin x + B cos x) for some consts. A and B , = "ex cosa giving y(x)= e"(c1 cosx + C2 sin x - ~ cosx) . (e) yc(x) = C1 sin x + c2 cosx for arb. consts. C1 and c2 , yp(x)=(Ax+ B) +x(Cx+ D) sinx+ x(Ex+ F) cosx , for some constants A, ... , F. Then yp(x) = x + ~ sin x sin2 - 20 COS x giving y(x) =C1 sinx + C2 cosx + x + x 4 sin x - r2 COS x . 4 2. Using the substitution y = vx to bring the homogeneous equations into separable form: (a) dv dx otady - =- e70. so that
Separating the variables x e dx dv _1 eº= - ln |x| + c . so that But y = vx so that ex =- ln |x|+c, for arbitrary constant c. (b) v + x- so that x dv 2v 02 -1 a du dx 2v 302 - 1 . Separating the variables: giving In |v2 - 1| = ln |x| + Inc say taking exponentials 12 - 1| = c|x | . But y = vx so that ly2 - x2| = c|x|3 , 2v dv v2 - 1 dx . 1 x for arbitrary constant c. (c) dv - = v + V1 + v2 dx dv so that x- = V1 + 22 . dx + du Separating the variables, arcsinhv= ln(v+v1+v2)=In|x|+Inc say taking exponentials v + v1 + v2 = c|x| . But y = vx so that y + Vx2 + y2 = cx|x| , for arbitrary constant c. Then fitting the boundary condition, c = 1 so that (x|x| - y)2 = x2 + y2 i.e. x4 - 2x|x|y = x2 giving y= x 2|x| -(x2 - 1) . Or: using arcsinh = In