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Question Number 131466 by Ahmed1hamouda last updated on 05/Feb/21

Commented by Ahmed1hamouda last updated on 05/Feb/21

$$$$$$\mathrm{solve}\mathrm{the}\mathrm{differential}\mathrm{equation}\mathrm{n})\mathrm{D}$$

Answered by EDWIN88 last updated on 05/Feb/21

$$\frac{dy}{dx}+\frac{y\ln x}{x}=y.e^x$$$$\frac{dy}{dx}=y(e^x-\frac{\ln x}{x})$$$$\frac{dy}{y}=(e^x-\frac{\ln x}{x})dx$$$$\ln y=e^x-\frac{1}{2}\ln {}^2\left(x\right)+c$$$$\ln y^2+{\left(\ln x\right)}^2=2e^x+C$$$$$$

Commented by Ahmed1hamouda last updated on 05/Feb/21

$$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}+\frac{\mathit{y}\mathit{l}\mathit{n}\left(\mathit{y}\right)}{\mathit{x}}=\mathit{y}.{\mathit{e}}^{\mathit{x}}$$$$\frac{1}{\mathit{y}}.\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{\mathit{l}\mathit{n}\left(\mathit{y}\right)}{\mathit{x}}={\mathit{e}}^{\mathit{x}}$$$$\mathit{l}\mathit{n}\left(\mathit{y}\right)=\mathit{u}\therefore \frac{1}{\mathit{y}}.\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{\mathit{d}\mathit{u}}{\mathit{d}\mathit{x}}$$$$\frac{\mathit{d}\mathit{u}}{\mathit{d}\mathit{x}}+\frac{\mathit{u}}{\mathit{x}}={\mathit{e}}^{\mathit{x}}\to \mathrm{I}.\mathrm{F}.={\mathit{e}}^{\int \frac{1}{\mathit{x}}\mathit{d}\mathit{x}}={\mathit{e}}^{\mathit{l}\mathit{n}\left(\mathit{x}\right)}=\mathit{x}$$$$\mathit{u}\mathit{x}=\int {\mathit{x}\mathit{e}}^{\mathit{x}}.\mathit{d}\mathit{x}\to \mathit{x}\mathit{l}\mathit{n}\left(\mathit{y}\right)={\mathit{x}\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{\mathit{x}}+\mathit{c}$$

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