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Question Number 90307 by niroj last updated on 22/Apr/20

$$\mathbf{Solve}\mathbf{the}\mathbf{differential}\mathbf{equation}.$$$$({\mathbf{x}}^2{\mathbf{D}}^2-2)\mathbf{y}={\mathbf{x}}^2+\frac{1}{\mathbf{x}}.$$

Answered by TANMAY PANACEA. last updated on 22/Apr/20

$$x^2\frac{d^{2}y}{{dx}^2}-2y=x^2+\frac{1}{x}$$$$x=e^t$$$$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=\frac{1}{e^t}\frac{dy}{dt}$$$$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{1}{e^t}\times \frac{dy}{dt}\right)\times \frac{dt}{dx}=\frac{1}{e^t}[\frac{1}{e^t}\times \frac{d^{2}y}{{dt}^2}-\frac{1}{e^t}\times \frac{dy}{dt}]$$$${\left(e^t\right)}^2\frac{d^{2}y}{{dx}^2}=\frac{d^{2}y}{{dt}^2}-\frac{dy}{dt}$$$$x^2\frac{d^{2}y}{{dx}^2}=\frac{d^{2}y}{{dt}^2}-\frac{dy}{dt}$$$$\mathit{s}\mathit{o}$$$$\frac{{\mathit{d}}^2\mathit{y}}{{\mathit{d}\mathit{t}}^2}-\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{t}}-2\mathit{y}={\mathit{e}}^{2\mathit{t}}+{\mathit{e}}^{-\mathit{t}}$$$$\mathit{l}\mathit{e}\mathit{t}\mathit{y}={\mathit{e}}^{\mathit{m}\mathit{t}}$$$$({\mathit{m}}^2-\mathit{m}-2){\mathit{e}}^{\mathit{m}\mathit{t}}=0\mathit{f}\mathit{o}\mathit{r}\mathit{C}.F$$$$(m-2)(m+1)=0$$$$C.F={Ae}^{2t}+{Be}^{-t}=\mathit{A}{x}^2+\frac{B}{x}$$$$P.I$$$$let\frac{d}{dt}=\theta$$$$y=\frac{e^{2t}+e^{-t}}{({\theta }^2-\theta -2)}=\frac{x^2+\frac{1}{x}}{(\theta -2)(\theta +1)}=\frac{1}{3}\times \frac{(\theta +1)-(\theta -2)}{(\theta -2)(\theta +1)}\times (x^2+\frac{1}{x})$$$$=\frac{1}{3}[\frac{1}{\theta -2}-\frac{1}{\theta +1}](x^2+\frac{1}{x})$$$$=\frac{1}{3}\times [x^2\int x^{-2-1}(x^2+\frac{1}{x})dx-x^{-1}\int x^{1-1}(x^2+\frac{1}{x})dx]$$$$=\frac{1}{3}[x^2\int (\frac{1}{x}+\frac{1}{x^4})dx-\frac{1}{x}\int (x^2+\frac{1}{x})dx]$$$$=\frac{1}{3}[x^2\times lnx+x^2\times \frac{1}{x^3\times (-3)}-\frac{1}{x}\times \frac{x^3}{3}-\frac{1}{x}lnx]$$$$=\frac{x^{2}lnx}{3}-\frac{1}{9}\times \frac{1}{x}-\frac{x^2}{9}-\frac{lnx}{3x}$$$$=(x^2-\frac{1}{x})\times \frac{lnx}{3}-\frac{1}{9}(x^2+\frac{1}{x})$$$$y={Ax}^2+\frac{B}{x}+\frac{x^{2}lnx}{3}-\frac{lnx}{3x}-\frac{x^2}{9}-\frac{1}{9x}$$$$y=C_1{x}^2+\frac{C_2}{x}+\frac{x^{2}lnx}{3}-\frac{lnx}{3x}$$$$$$$$$$$$$$$$$$

Commented by niroj last updated on 22/Apr/20

$$\mathrm{I}\mathrm{apriciate}\mathrm{your}\mathrm{trying}\mathrm{effort}\mathrm{but}$$$$\mathrm{Answer}\mathrm{should}\mathrm{make}\mathrm{sure}$$$$\mathbf{y}={\mathbf{C}}_1{\mathrm{x}}^{-1}+{\mathrm{C}}_2{\mathbf{x}}^2+\frac{1}{3}{\mathbf{x}}^2\mathbf{log}\mathbf{x}-\frac{1}{3\mathbf{x}}\mathbf{log}\mathbf{x}.$$

Commented by TANMAY PANACEA. last updated on 22/Apr/20

$$ihavecorrected...usingDanielandmurrydiff$$$$calbook$$

Commented by niroj last updated on 23/Apr/20

$$\mathrm{now}\mathrm{done}\mathrm{dear}.$$

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