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Question Number 83849 by Rio Michael last updated on 06/Mar/20

$$\mathrm{Gven}\mathrm{that}y=e^{-x}\sin bx,\mathrm{where}b\mathrm{is}\mathrm{a}\mathrm{constant},\mathrm{show}\mathrm{that}$$$$\frac{d^{2}y}{{dx}^2}+2\frac{dy}{dx}+(1+b^2)y=0.$$

Commented by niroj last updated on 06/Mar/20

$$$$$$\mathrm{Given},$$$$\mathrm{y}={\mathrm{e}}^{-\mathrm{x}}\sin \mathrm{bx}....\left(\mathrm{i}\right)$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-1.{\mathrm{e}}^{-\mathrm{x}}.\mathrm{sinbx}+{\mathrm{e}}^{-\mathrm{x}}.\mathrm{b}.\cos \mathrm{bx}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}+{\mathrm{e}}^{-\mathrm{x}}.\mathrm{bcosbx}.....\left(\mathrm{ii}\right)(\because \mathrm{y}={\mathrm{e}}^{-\mathrm{x}}\sin \mathrm{bx})$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-\frac{\mathrm{dy}}{\mathrm{dx}}+(-1.{\mathrm{e}}^{-\mathrm{x}}\mathrm{b}\cos \mathrm{bx}-{\mathrm{e}}^{-\mathrm{x}}{\mathrm{b}}^2\mathrm{sinbx})$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-\frac{\mathrm{dy}}{\mathrm{dx}}-{\mathrm{e}}^{\mathrm{x}}\mathrm{bcosbx}-{\mathrm{b}}^2\mathrm{y}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-\frac{\mathrm{dy}}{\mathrm{dx}}-(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y})-{\mathrm{b}}^2\mathrm{y}\{\mathrm{from}..\left(\mathrm{ii}\right)\}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}-{\mathrm{b}}^2\mathrm{y}$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=-2\frac{\mathrm{dy}}{\mathrm{dx}}-(1+{\mathrm{b}}^2)\mathrm{y}$$$$\therefore \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}+2\frac{\mathrm{dy}}{\mathrm{dx}}+(1+{\mathrm{b}}^2)\mathrm{y}=0\mathrm{hence}\mathrm{proved}//.$$

Commented by Rio Michael last updated on 06/Mar/20

$$thankssir$$

Commented by niroj last updated on 07/Mar/20

$$youmustwelcomesir.$$

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