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Question Number 77483 by aliesam last updated on 06/Jan/20

$$solve$$$$y^{\prime }cos\left(x\right)+\frac{1}{2}ysin\left(x\right)=e^x\sqrt{sin\left(x\right)}$$

Commented by mathmax by abdo last updated on 06/Jan/20

$$\left(he\right)\to cosx)y^{{}^{\prime }}+\frac{sinx}{2}y=0\Rightarrow \left(cosx\right)y^{{}^{\prime }}=-\frac{sinx}{2}y\Rightarrow$$$$\frac{y^{{}^{\prime }}}{y}=-\frac{1}{2}tanx\Rightarrow ln\mid y\mid =\frac{1}{2}\times \int \frac{-sinx}{cosx}dx+\alpha \Rightarrow$$$$ln\mid y\mid =\frac{1}{2}ln\mid cosx\mid +\alpha \Rightarrow y\left(x\right)=K\sqrt{\mid cosx\mid }$$$$letdeterminethesolutionon\mathrm{\Omega }=\{x/cosx>0\}$$$$mvcmethody\left(x\right)=K\left(x\right)\sqrt{cosx}\Rightarrow y^{{}^{\prime }}=K^{{}^{\prime }}\sqrt{cosx}+K\times \frac{-sinx}{2\sqrt{cosx}}$$$$\left(e\right)\Rightarrow K^{{}^{\prime }}cosx\sqrt{cosx}-\frac{1}{2}K\sqrt{cosx}sinx+\frac{1}{2}K\sqrt{cosx}sinx=e^x\sqrt{sinx}$$$$\Rightarrow K^{{}^{\prime }}=\frac{e^x\sqrt{sinx}}{cosx\sqrt{cosx}}\Rightarrow K\left(x\right)={\int }_0^x\frac{e^t\sqrt{sint}}{{\left(cost\right)}^{\frac{3}{2}}}dt+c\Rightarrow$$$$y\left(x\right)=\sqrt{cosx}\{{\int }_0^x\frac{e^t\sqrt{sint}}{{\left(cost\right)}^{\frac{3}{2}}}dt+C)$$

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