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Question Number 85165 by mathmax by abdo last updated on 19/Mar/20

$$sove\left({sin}^{2}x\right)y^{{}^{\prime }}+\left(cosx\right)y=x$$

Commented by mathmax by abdo last updated on 20/Mar/20

$$\left(he\right)\to \left({sin}^{2}x\right)y^{{}^{\prime }}+\left(cosx\right)y=0\Rightarrow \frac{y^{{}^{\prime }}}{y}=-\frac{cosx}{{sin}^{2}x}\Rightarrow$$$$ln\mid y\mid =\frac{1}{sinx}+k\Rightarrow y\left(x\right)=C{e}^{\frac{1}{sinx}}letusemvcmethod$$$$y^{{}^{\prime }}\left(x\right)=C^{{}^{\prime }}{e}^{\frac{1}{sinx}}+C(-\frac{cosx}{{sin}^{2}x})e^{\frac{1}{sinx}}$$$$\left(e\right)\Rightarrow C^{{}^{\prime }}{sin}^{2}x{e}^{\frac{1}{sinx}}-Ccosx{e}^{\frac{1}{sinx}}+Ccosx{e}^{\frac{1}{sinx}}=x\Rightarrow$$$$C{}^{{}^{\prime }}=\frac{x}{{sin}^{2}x}{e}^{-\frac{1}{sinx}}\Rightarrow C\left(x\right)={\int }^x\frac{u}{{sin}^{2}u}{e}^{-\frac{1}{sinu}}du+\lambda \Rightarrow$$$$y\left(x\right)=({\int }^x\frac{u}{{sin}^{2}u}{e}^{-\frac{1}{sinu}}du+\lambda )e^{\frac{1}{sinx}}isthegeneral[solution$$

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