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Question Number 128432 by bemath last updated on 07/Jan/21

$$({3\mathrm{x}}^2+1){\mathrm{yy}}^{\prime }+{3\mathrm{xy}}^2=\cos \mathrm{x}.{\mathrm{e}}^{\sin \mathrm{x}}$$

Answered by liberty last updated on 07/Jan/21

$$\mathrm{let}{\mathrm{y}}^2=\mathrm{v}\left(\mathrm{x}\right)\Rightarrow {2\mathrm{yy}}^{\prime }=\frac{\mathrm{dv}}{\mathrm{dx}}$$$$\Rightarrow {\mathrm{yy}}^{\prime }=\frac{1}{2}\frac{\mathrm{dv}}{\mathrm{dx}};\frac{1}{2}({3\mathrm{x}}^2+1)\frac{\mathrm{dv}}{\mathrm{dx}}+3\mathrm{vx}=\cos \mathrm{x}.{\mathrm{e}}^{\sin \mathrm{x}}$$$$\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}+\frac{6\mathrm{vx}}{{3\mathrm{x}}^2+1}=\frac{2\cos \mathrm{x}.{\mathrm{e}}^{\sin \mathrm{x}}}{{3\mathrm{x}}^2+1}$$$$\mathrm{put}\mathrm{integrating}\mathrm{factor}\mu ={\mathrm{e}}^{\int \frac{6\mathrm{x}}{{3\mathrm{x}}^2+1}\mathrm{dx}}={\mathrm{e}}^{\ln ({3\mathrm{x}}^2+1)}={3\mathrm{x}}^2+1$$$$\mathrm{we}\mathrm{get}\frac{\mathrm{d}}{\mathrm{dx}}\left(({3\mathrm{x}}^2+1)\mathrm{v}\right)=2\cos \mathrm{x}.{\mathrm{e}}^{\sin \mathrm{x}}$$$$\Rightarrow ({3\mathrm{x}}^2+1)\mathrm{v}=\int 2\cos \mathrm{x}.{\mathrm{e}}^{\sin \mathrm{x}}\mathrm{dx}$$$$\Rightarrow ({3\mathrm{x}}^2+1)\mathrm{v}={2\mathrm{e}}^{\sin \mathrm{x}}+\mathrm{C}$$$$\Rightarrow \mathrm{v}=\frac{{2\mathrm{e}}^{\sin \mathrm{x}}+\mathrm{C}}{{3\mathrm{x}}^2+1};{\mathrm{y}}^2=\frac{{2\mathrm{e}}^{\sin \mathrm{x}}+\mathrm{C}}{{3\mathrm{x}}^2+1}$$

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