homework 09 solutions

# homework 09 solutions - Section 3.5 7 First we substitute...

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Unformatted text preview: Section 3.5 7. First we substitute sinh x = (e" — e‘ )/ 2 on the right-hand side of the differential equation. Then: yum = Ae’+Be"‘; —3A=l/2, —3B=-—1/2; yp = (e‘x—e")/6 = —(1/3)sinhx 18. First we note the duplication with the complementary function y0 = cle" + cze‘ + c3e'2" + c4e2". Then: ytrial = we") +x(B + Cx) e2”; —6A =1, 123+3sc = 0, 24C = —1; yp = —(24xe" — 19xe2" + 6x2e7")/ 144 28. y6 = (c1+c2x)+(c3cos3x+c3 sin3x) yi = (A+Bx+Cx2)cos3x+(D+Ex+Fx2)sin3x yp = x-[(A+Bx+Cx2)cos3x+(D+Ex+Fx2)sin3x] 36. y0 = cl+c2x+c3e'2"+c4ez"; ytlr = xz-(A+Bx+Cx2) yg = cl+c2x+c3e"2"+c4ezx-x2/16—x4/48 c1+c3+c4=1, c2—2c3+2c4=1, 4c3+4c4—1/8=—1, —803+8c4=—1 y(x) = (234 + 240x — 912-2" — 33122" ~12x2 — 4x4 ) / 192 43. (a) cos 3x + i sin 3x = (cosx + isin x)3 = cos3 x+3icos2 xsinx-3cosxsin2 x-—z'sin3 x When we equate real parts we get the equation cos3 x — 3(cosx)(l —cos2 x) = 40053 x — 3cosx and readily solve for cos3 x = %cosx + %cos 3x. The formula for sin3 x is derived similarly by equating imaginary parts in the ﬁrst equation above. (b) Upon substituting the trial solution yp = A cos x + B sin x + C cos 3x + D sin 3x in the differential equation y" + 4y = %cosx + %cos 3x, we ﬁnd that A = 1/4, B = 0, C = —l/20, D = 0. The resulting general solution is y(x) = clcos 2x + czsin 2x + (l/4)cos x — (1/20)cos 3x. 63. This is simply a matter of solving the equations in (31) for the derivatives u, _ _y2(x)f(x) and u, _ y1(x)f(x) ‘ W(x) 2 W<x) ’ integrating each, and then substituting the results in (32). 64. Here we have y1(x) = cos x, y2(x) = sin x, W(x) = 1, f(x) = 2sinx , so (33) gives yp(x) = — (cosx) Isinx - 23inxdx +(sinx) Icosx - 25inxdx = — (cos x) [(1— cos 2x) dx +(sin x) I2(sin x) - cos x dx = — (cos x)(x — sin x cos x) + (sin x)(sin2 x) = - x cosx + (sin x)(cos2 x + sin2 x) yp(x) = —xcosx+sinx But we can drop the term sin x because it satisﬁes the associated homogeneous equation y" + y = 0. ...
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