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Question Number 37344 by math khazana by abdo last updated on 12/Jun/18

$$solvethed.e.y^{{}^{\prime }}-xy=cosx.$$

Commented by math khazana by abdo last updated on 12/Jun/18

$$he\Rightarrow y^{{}^{\prime }}-xy=0\Rightarrow \frac{y^{{}^{\prime }}}{y}=x\Rightarrow ln\mid y\mid =\frac{x^2}{2}+c\Rightarrow$$$$y=k{e}^{\frac{x^2}{2}}mvcmethodgive$$$$y^{{}^{\prime }}=k^{{}^{\prime }}{e}^{\frac{x^2}{2}}+kx{e}^{\frac{x^2}{2}}\left(e\right)\Rightarrow k^{{}^{\prime }}{e}^{\frac{x^2}{2}}+kx{e}^{\frac{x^2}{2}}-xk{e}^{\frac{x^2}{2}}=cosx$$$$\Rightarrow k^{{}^{\prime }}=e^{-\frac{x^2}{2}}cosx\Rightarrow k\left(x\right)={\int }_{.}^x{e}^{-\frac{t^2}{2}}cost+\lambda \Rightarrow$$$$y\left(x\right)=({\int }_{.}^x{e}^{-\frac{t^2}{2}}costdt+\lambda )e^{\frac{x^2}{2}}\Rightarrow$$$$y\left(x\right)=e^{\frac{x^2}{2}}{\int }_{.}^x{e}^{-\frac{t^2}{2}}costdt+\lambda e^{\frac{x^2}{2}}.$$

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