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Question Number 96138 by john santu last updated on 30/May/20

$${xy}^{\prime }+y^2=x^2{e}^x$$

Commented by Sourav mridha last updated on 30/May/20

$${\mathit{i}\mathit{t}}^{\prime }\mathrm{s}\mathrm{wrong},\mathrm{it}\mathit{s}\mathit{h}\mathit{o}\mathit{u}\mathit{l}\mathit{d}\mathit{b}\mathit{e}$$$${\mathit{x}\mathit{y}}^{{}^{\prime }}+\mathit{y}={\mathit{x}}^2{\mathit{e}}^{\mathit{x}}.$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathit{a}\mathit{n}\mathit{s}\mathit{i}\mathit{s}--$$$$\mathit{x}\mathit{y}={\mathit{x}}^2{\mathit{e}}^{\mathit{x}}-2[{\mathit{x}\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{\mathit{x}}]+\mathit{c}$$

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