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Question Number 133961 by Ahmed1hamouda last updated on 25/Feb/21

Answered by mathmax by abdo last updated on 27/Feb/21

$${\mathrm{y}}^{{}^{\prime }}+\mathrm{sinx}\mathrm{y}=-{\mathrm{x}}^3$$$$\mathrm{h}\to {\mathrm{y}}^{{}^{\prime }}+\mathrm{sinxy}=0\Rightarrow {\mathrm{y}}^{{}^{\prime }}=-\mathrm{sinxy}\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=-\mathrm{sinx}\Rightarrow \ln \mid \mathrm{y}\mid =\mathrm{cosx}+\mathrm{c}\Rightarrow$$$$\mathrm{y}=\mathrm{k}{\mathrm{e}}^{\mathrm{cosx}}\mathrm{lagrange}\mathrm{method}\to {\mathrm{y}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{\mathrm{cosx}}-\mathrm{sinx}{\mathrm{ke}}^{\mathrm{cosx}}$$$$\mathrm{e}\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{\mathrm{cosx}}-\mathrm{sinx}{\mathrm{ke}}^{\mathrm{cosx}}+\mathrm{sinxk}{\mathrm{e}}^{\mathrm{cosx}}=-{\mathrm{x}}^3\Rightarrow$$$${\mathrm{k}}^{{}^{\prime }}=-{\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{cosx}}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)=-\int {\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{cosx}}\mathrm{dx}+\lambda \Rightarrow$$$$\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{is}\mathrm{y}\left(\mathrm{x}\right)=(\lambda -\int {\mathrm{x}}^3{\mathrm{e}}^{-\mathrm{cosx}}\mathrm{dx}){\mathrm{e}}^{\mathrm{cosx}}$$

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