Chem Differential Eq HW Solutions Fall 2011 178

# Chem Differential Eq HW Solutions Fall 2011 178 - A178...

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Unformatted text preview: A178 Appendix A Ordinary Diﬀerential Equations: Review of Concepts and Methods From Exercise 25, yh = c1ex + c2 e3x . For a particular solution try yp = Axe3x + Bxex . 49. y − 3 y + 2 y = 3x4ex + x e−2x cos 3x. Characteristic equation λ2 − 3 λ + 2 = 0 ⇒ λ1 = 1, λ2 = 2. So yh = c1ex + c2 e2x . For a particular solution, try yp = x(Ax4 + Bx3 + Cx2 + Dx + E )ex +(Gx + H ) e−2x cos 3x +(Kx + L) e−2x sin 3x. 53. y − 2 y + y = 6x − ex . Characteristic equation λ2 − 2 λ + 1 = 0 ⇒ λ1 = 1(double root). So yh = c1ex + c2 xex . For a particular solution, try yp = Ax + B + Cx2ex . 57. y + 4y = cos ωx. We have yh = c1 cos 2x + c2 sin 2x. If ω = ±2, a particular solution of y + 4y = cos ωx is yp = A cos ωx. So yp = −Aω2 cos ωx. Plugging into the equation, we ﬁnd A cos ωx(4 − ω2 ) = A(4 − ω2 ) = A= cos ωx; 1; 1 . 4 − ω2 Note that 4 − ω2 = 0, because ω = ±2. So the general solution in this case is of the form cos ωx yg = c1 cos 2x + c2 sin 2x + . 4 − ω2 If ω = ±2, then we modify the particular solution and use yp = x A cos ωx + B sin ωx . Then yp = yp = A cos ωx + B sin ωx + xω − A sin ωx + B cos ωx , xω2 − A cos ωx − B sin ωx + 2ω − A sin ωx + B cos ωx . Plug into the left side of the equation xω2 − A cos ωx − B sin ωx + 2ω − A sin ωx + B cos ωx + 4x A cos ωx + B sin ωx . Using ω2 = 4, this becomes 2ω − A sin ωx + B cos ωx . This should equal cos ωx. So A = 0 and 2ωB = 1 or B = 1/(2ω). yg = c1 cos 2x + c2 sin 2x + 1 x sin ωx (ω = ±2). 2ω Note that if ω = 2 or ω = −2, the solution is 1 yg = c1 cos 2x + c2 sin 2x + x sin 2x. 4 ...
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