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Question Number 96140 by Farruxjano last updated on 30/May/20

$${\mathit{x}\mathit{y}}^{\prime }+{\mathit{y}}^2={\mathit{x}}^2{\mathit{e}}^{\mathit{x}}\Rightarrow {\mathit{y}}^{\prime }={\mathit{x}\mathit{e}}^{\mathit{x}}-\frac{{\mathit{y}}^2}{\mathit{x}}\Rightarrow$$$$\mathit{y}={\mathit{x}\mathit{y}\mathit{e}}^{\mathit{x}}-\frac{1}{3\mathit{x}}\cdot {\mathit{y}}^3\Rightarrow \frac{{\mathit{y}}^3}{3\mathit{x}}+\mathit{y}-{\mathit{x}\mathit{y}\mathit{e}}^{\mathit{x}}=0$$$$\mathit{y}(\frac{{\mathit{y}}^2}{3\mathit{x}}+1-{\mathit{x}\mathit{e}}^{\mathit{x}})=0\Rightarrow \mathit{y}=\pm \sqrt{3\mathit{x}({\mathit{x}\mathit{e}}^{\mathit{x}}-1)}$$

Commented by john santu last updated on 30/May/20

$$\mathrm{how}\mathrm{to}\mathrm{get}$$$${\mathrm{y}}^{\prime }={xe}^x-\frac{y^2}{x}\Rightarrow \frac{dy}{dx}={xe}^x-\frac{y^2}{x}$$$$\int dy=\int ({xe}^x-\frac{y^2}{x})dx$$$$y={xye}^x-\frac{y^3}{3x}???$$

Commented by mr W last updated on 30/May/20

$$pleasepostyouranswerinthesame$$$$threadasthequestionitself.{don}^{\prime }t$$$$openanewone!$$

Commented by mr W last updated on 30/May/20

$$besides,yoursolutioniswrong.$$

Commented by Farruxjano last updated on 30/May/20

$$\mathrm{I}\mathrm{tried},\mathrm{I}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{how}\mathrm{to}\mathrm{solve}$$$$\mathrm{but}\mathrm{i}\mathrm{was}\mathrm{going}\mathrm{to}\mathrm{know}\mathrm{the}\mathrm{solution}$$

Commented by mr W last updated on 30/May/20

$$itisabsolutelyokwhenyou{don}^{\prime }t$$$$knowhowtosolveaquestion.ialso$$$${don}^{\prime }tknowhowtosolveit.theni$$$$keepbeingsilentandlookhowthe$$$$otherssolveitandlearnfromthem.$$

Answered by john santu last updated on 30/May/20

$$\mathrm{it}\mathrm{impossible}$$

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