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Question Number 95695 by mathmax by abdo last updated on 27/May/20

$$\mathrm{solve}(\mathrm{x}+1){\mathrm{y}}^{{}^{\prime }}-{\mathrm{x}}^3\mathrm{y}=\arctan \left(2\mathrm{x}\right)$$

Answered by mathmax by abdo last updated on 27/May/20

$$\left(\mathrm{he}\right)\Rightarrow (\mathrm{x}+1){\mathrm{y}}^{{}^{\prime }}-{\mathrm{x}}^3\mathrm{y}=0\Rightarrow \frac{{\mathrm{y}}^{{}^{\prime }}}{\mathrm{y}}=\frac{{\mathrm{x}}^3}{\mathrm{x}+1}\Rightarrow \ln \mid \mathrm{y}\mid =\int \frac{{\mathrm{x}}^3}{\mathrm{x}+1}\mathrm{dx}+\mathrm{c}$$$$=\int \frac{{\mathrm{x}}^3+1-1}{\mathrm{x}+1}\mathrm{dx}+\mathrm{c}=\int \frac{{\mathrm{x}}^3+1}{\mathrm{x}+1}\mathrm{dx}-\int \frac{\mathrm{dx}}{\mathrm{x}+1}+\mathrm{c}=\int ({\mathrm{x}}^2-\mathrm{x}+1)\mathrm{dx}-\ln \mid \mathrm{x}+1\mid +\mathrm{c}$$$$=\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}-\ln \mid \mathrm{x}+1\mid +\mathrm{c}\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{\mid \mathrm{x}+1\mid }{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}\mathrm{let}\mathrm{find}\mathrm{the}\mathrm{solution}\mathrm{on}$$$$]-1,+\mathrm{\infty }[\Rightarrow \mathrm{y}=\frac{\mathrm{k}}{\mathrm{x}+1}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}\mathrm{mvc}\mathrm{method}\to$$$${\mathrm{y}}^{{}^{\prime }}=\frac{{\mathrm{k}}^{{}^{\prime }}}{\mathrm{x}+1}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}+\mathrm{k}(-\frac{1}{{(\mathrm{x}+1)}^2}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}+\frac{1}{\mathrm{x}+1}\times ({\mathrm{x}}^2-\mathrm{x}+1){\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}})$$$$\left(\mathrm{e}\right)\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}-\frac{\mathrm{k}}{(\mathrm{x}+1)}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}+\mathrm{k}({\mathrm{x}}^2-\mathrm{x}+1){\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}-\frac{{\mathrm{kx}}^3}{\mathrm{x}+1}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}=\arctan \left(2\mathrm{x}\right)$$$$\Rightarrow {\mathrm{k}}^{{}^{\prime }}=\arctan \left(2\mathrm{x}\right){\mathrm{e}}^{-\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}\Rightarrow \mathrm{k}\left(\mathrm{x}\right)={\int }_{.}^{\mathrm{x}}\arctan \left(2\mathrm{u}\right){\mathrm{e}}^{-\frac{{\mathrm{u}}^3}{3}-\frac{{\mathrm{u}}^2}{2}+\mathrm{u}}\mathrm{du}+\mathrm{c}\Rightarrow$$$$\mathrm{y}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}+1}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}({\int }_{.}^{\mathrm{x}}\arctan \left(2\mathrm{u}\right){\mathrm{e}}^{-\frac{{\mathrm{u}}^3}{3}-\frac{{\mathrm{u}}^2}{2}+\mathrm{u}}\mathrm{du}+\mathrm{c})$$$$=\frac{\mathrm{c}}{\mathrm{x}+1}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}+\frac{1}{\mathrm{x}+1}{\mathrm{e}}^{\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^2}{2}+\mathrm{x}}{\int }_{.}^{\mathrm{x}}\arctan \left(2\mathrm{u}\right){\mathrm{e}}^{-\frac{{\mathrm{u}}^3}{3}-\frac{{\mathrm{u}}^2}{2}+\mathrm{u}}\mathrm{du}$$

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