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(D2-2D+5)y = e2x sin x |
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Answer» (D2 - 2D + 5)y = e2x sinx For C.F, use m in place of D m2 - 2m + 5 = 0 ⇒ m = \(\frac{2 ± \sqrt{4-20}}{2}\) = \(\frac{2±4c}{2}\) = 1 ± 2c \(\therefore\) C.F = ex (C1cos 2x + c2 sin 2x) P.I = \(\frac{1}{D^2 - 2D +5}\) e2x sinx = e2x \(\frac{1}{(D+2)^2 -2(D+2) + 5} sinx\) \(\bigg(\therefore \frac{1}{f(D) }e^{ax}V = e^{ax} \frac{1}{f(D+a)V}\bigg)\) = e2x \(\frac{1}{D^2+2D + 5} sinx\) = e2x \(\frac{1}{-1+2d+5}sinx\) \(\bigg(\therefore \frac{1}{f(D^2)} sinax = \frac{1}{f(-a^2)} sinax\bigg)\) = e2x \(\frac{1}{2D + 4}\) sinx = \(\frac{e^{2x}}{2} \frac{D-2}{D^2 -4} sinx\) = \(\frac{e^{2x}}{2}\) \(\frac{(D-2)sinx}{-1-4}\) = \(\frac{e^{2x}}{-10}(Dsinx - 2sinx)\) = \(\frac{-e^{2x}}{10}(cos x -2sinx)\) Complete solution y = ex (c1cos 2x + c2 sin2x) = \(\frac{-e^{2x}}{10}\) (cosx -2sinx) |
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