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Question Number 77483 by aliesam last updated on 06/Jan/20
solve   y′ cos(x) +(1/2) y sin(x) = e^x  (√(sin(x)))
$${solve}\: \\ $$$${y}'\:{cos}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:{y}\:{sin}\left({x}\right)\:=\:{e}^{{x}} \:\sqrt{{sin}\left({x}\right)}\:\: \\ $$
Commented by mathmax by abdo last updated on 06/Jan/20
(he)→cosx)y^′  +((sinx)/2)y=0 ⇒(cosx)y^′ =−((sinx)/2)y ⇒  (y^′ /y) =−(1/2)tanx ⇒ln∣y∣=(1/2)×∫((−sinx)/(cosx))dx +α ⇒  ln∣y∣=(1/2)ln∣cosx∣ +α ⇒y(x)=K(√(∣cosx∣))  let determine the solution on Ω={x/cosx>0}  mvc method  y(x)=K(x)(√(cosx)) ⇒y^′ =K^′ (√(cosx))+K×((−sinx)/(2(√(cosx))))  (e) ⇒K^′ cosx(√(cosx))−(1/2)K(√(cosx))sinx +(1/2)K(√(cosx))sinx =e^x (√(sinx))  ⇒K^′ =((e^x (√(sinx)))/(cosx(√(cosx)))) ⇒K(x)=∫_0 ^x  ((e^t (√(sint)))/((cost)^(3/2) ))dt +c ⇒  y(x)=(√(cosx)){ ∫_0 ^x  ((e^t (√(sint)))/((cost)^(3/2) ))dt +C)
$$\left.\left({he}\right)\rightarrow{cosx}\right){y}^{'} \:+\frac{{sinx}}{\mathrm{2}}{y}=\mathrm{0}\:\Rightarrow\left({cosx}\right){y}^{'} =−\frac{{sinx}}{\mathrm{2}}{y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=−\frac{\mathrm{1}}{\mathrm{2}}{tanx}\:\Rightarrow{ln}\mid{y}\mid=\frac{\mathrm{1}}{\mathrm{2}}×\int\frac{−{sinx}}{{cosx}}{dx}\:+\alpha\:\Rightarrow \\ $$$${ln}\mid{y}\mid=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{cosx}\mid\:+\alpha\:\Rightarrow{y}\left({x}\right)={K}\sqrt{\mid{cosx}\mid} \\ $$$${let}\:{determine}\:{the}\:{solution}\:{on}\:\Omega=\left\{{x}/{cosx}>\mathrm{0}\right\} \\ $$$${mvc}\:{method}\:\:{y}\left({x}\right)={K}\left({x}\right)\sqrt{{cosx}}\:\Rightarrow{y}^{'} ={K}^{'} \sqrt{{cosx}}+{K}×\frac{−{sinx}}{\mathrm{2}\sqrt{{cosx}}} \\ $$$$\left({e}\right)\:\Rightarrow{K}^{'} {cosx}\sqrt{{cosx}}−\frac{\mathrm{1}}{\mathrm{2}}{K}\sqrt{{cosx}}{sinx}\:+\frac{\mathrm{1}}{\mathrm{2}}{K}\sqrt{{cosx}}{sinx}\:={e}^{{x}} \sqrt{{sinx}} \\ $$$$\Rightarrow{K}^{'} =\frac{{e}^{{x}} \sqrt{{sinx}}}{{cosx}\sqrt{{cosx}}}\:\Rightarrow{K}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:\frac{{e}^{{t}} \sqrt{{sint}}}{\left({cost}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt}\:+{c}\:\Rightarrow \\ $$$${y}\left({x}\right)=\sqrt{{cosx}}\left\{\:\int_{\mathrm{0}} ^{{x}} \:\frac{{e}^{{t}} \sqrt{{sint}}}{\left({cost}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dt}\:+{C}\right) \\ $$

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