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Question Number 86307 by niroj last updated on 28/Mar/20

$$\mathbf{Solve}\mathbf{the}\mathbf{following}\mathbf{equation}:$$$$\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}-2\frac{\mathbf{dy}}{\mathbf{dx}}+\mathbf{y}=\mathbf{x}{\mathbf{e}}^{\mathbf{x}}\mathbf{sin}\mathbf{x}.$$

Answered by TANMAY PANACEA. last updated on 28/Mar/20

$$y=e^{mx}$$$$(m^2-2m+1)e^{mx}=0$$$$e^{mx}\ne 0m=1,1$$$$C.F={Ae}^x+{Bxe}^x$$$$P.I=\frac{{xe}^{x}sinx}{D^2-2D+1}$$$$=e^x.\frac{xsinx}{{(D+1)}^2-2(D+1)+1}$$$$=e^x.\frac{xsinx}{D^2+2D+1-2D-2+1}$$$$=e^x.\frac{xsinx}{D^2}$$$$now\frac{f\left(x\right)}{D}=\int f\left(x\right)dx$$$$\int xsinxdx$$$$x\int sinxdx-\int [\frac{dx}{dx}\int sinxdx]dx$$$$-xcosx+\int cosxdx$$$$=-xcosx+sinx+c$$$$\int (-xcosx+sinx+c)dx\to \frac{xsinx}{D^2}$$$$-x\int cosxdx+\int [\frac{dx}{dx}\int cosxdx]dx-cosx+cx+c_1$$$$-xsinx-sinx-cosx+cx+c_1$$$$y={Ae}^x+{Bxe}^x-(1+x)sinx-cosx+cx+c_1$$

Commented by niroj last updated on 28/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{your}\mathrm{effort}\mathrm{Sir},$$$$\mathrm{but}\mathrm{answer}\mathrm{should}\mathrm{be}$$$$({\mathrm{C}}_1+{\mathrm{C}}_2\mathrm{x}){\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}(\mathrm{x}\sin \mathrm{x}+2\cos \mathrm{x}).$$

Commented by TANMAY PANACEA. last updated on 28/Mar/20

$$letmecheck$$

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