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Question Number 91825 by john santu last updated on 03/May/20

$$\left(\cos x\right)\frac{dy}{dx}-y\left(\sin x\right)=\cot \left(x\right)$$

Commented by Tony Lin last updated on 03/May/20

$${\left(ycosx\right)}^{\prime }=cotx$$$$ycosx=\int cotxdx=ln\mid sinx\mid +c$$$$y=\frac{ln\mid sinx\mid +c}{cosx}$$

Commented by john santu last updated on 03/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by mathmax by abdo last updated on 04/May/20

$$cosx{y}^{{}^{\prime }}-sinxy=cotanx$$$$\left(he\right)\to cosx{y}^{{}^{\prime }}-sinxy=0\Rightarrow cosx{y}^{{}^{\prime }}=sinxy\Rightarrow$$$$\frac{y^{{}^{\prime }}}{y}=\frac{sinx}{cosx}\Rightarrow ln\mid y\mid =-ln\left(cosx\right)+c\Rightarrow y=\frac{k}{cosx}$$$$letusemvcmethod\to y^{{}^{\prime }}=\frac{k^{{}^{\prime }}}{cosx}+k\frac{sinx}{{cos}^{2}x}$$$$\left(e\right)\Rightarrow k^{{}^{\prime }}+k\frac{sinx}{cosx}-\frac{ksinx}{cosx}=\frac{1}{tanx}\Rightarrow k^{{}^{\prime }}=\frac{cosx}{sinx}\Rightarrow$$$$k=\int \frac{cosx}{sinx}dx+c\Rightarrow k\left(x\right)=ln\left(sinx\right)+c\Rightarrow$$$$thegeneralsolutionisy\left(x\right)=\frac{1}{cosx}(ln\left(sinx\right)+c)$$$$=\frac{ln\left(sinx\right)}{cosx}+\frac{c}{cosx}$$

Answered by john santu last updated on 03/May/20

Commented by john santu last updated on 03/May/20

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Answered by niroj last updated on 03/May/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}-\left(\tan \mathrm{x}\right)\mathrm{y}=\mathrm{cosec}\mathrm{x}$$$$\mathrm{IF}={\mathrm{e}}^{\int \mathrm{Pdx}}={\mathrm{e}}^{-\int \tan \mathrm{xdx}}$$$$={\mathrm{e}}^{-\log \sec \mathrm{x}}={\mathrm{e}}^{\log \frac{1}{\sec \mathrm{x}}}$$$$\mathrm{IF}=\cos \mathrm{x}$$$$\mathrm{y}.\cos \mathrm{x}=\int \cos \mathrm{x}.\mathrm{cosec}\mathrm{x}\mathrm{dx}+\mathrm{c}$$$$\mathrm{y}\cos \mathrm{x}=\int \cos \mathrm{x}.\frac{1}{\sin \mathrm{x}}\mathrm{dx}+\mathrm{c}$$$$\mathrm{y}\cos \mathrm{x}=\int \cot \mathrm{x}\mathrm{dx}+\mathrm{c}$$$$\mathrm{y}\cos \mathrm{x}=\log \sin \mathrm{x}+\mathrm{c}$$$$\mathrm{y}=\frac{1}{\cos \mathrm{x}}(\log \sin \mathrm{x}+\mathrm{C})//.$$$$$$

Commented by john santu last updated on 03/May/20

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