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Miscellaneous Exercise 6(II) · Q117

Q.Verify: y=eaxsin⁡bxy=e^{ax}\sin bx, and d2ydx2−2adydx+(a2+b2)y=0\dfrac{d^2y}{dx^2}-2a\dfrac{dy}{dx}+(a^2+b^2)y=0

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y=eaxsin⁡bxy=e^{ax}\sin bx has characteristic roots a±bia\pm bi, so it solves the standard second-order equation d2ydx2−2adydx+(a2+b2)y=0\dfrac{d^2y}{dx^2}-2a\dfrac{dy}{dx}+(a^2+b^2)y=0. Directly: y′=eax(asin⁡bx+bcos⁡bx)y'=e^{ax}(a\sin bx+b\cos bx), y′′=eax[(a2−b2)sin⁡bx+2abcos⁡bx]y''=e^{ax}\left[(a^2-b^2)\sin bx+2ab\cos bx\right], and substituting confirms e …

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