# Solve differential equation \frac{dy}{dx}-(\cot x)y= \sin^3x

Solve differential equation $\frac{dy}{dx}-\left(\mathrm{cot}x\right)y={\mathrm{sin}}^{3}x$
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$\frac{dy}{dx}+P\left(n\right)y=Q\left(n\right)$
$I.F={e}^{\int p\left(x\right)dx}$
$={e}^{\int -\mathrm{cot}xdx}$
$={e}^{-\mathrm{ln}\mathrm{sin}n}$
$={e}^{\mathrm{ln}\frac{1}{\mathrm{sin}n}}$
$I.F.=\frac{1}{\mathrm{sin}n}$
$y\left(I.F.\right)=\int \left(I.F.\right)Q\left(x\right)$
$\frac{1}{\mathrm{sin}x}y=\int \left\{{\mathrm{sin}}^{3}x\right\}\left\{\mathrm{sin}x\right\}dn$
$\frac{y}{\mathrm{sin}x}=\int {\mathrm{sin}}^{2}ndx$
$\frac{y}{\mathrm{sin}x}=\int \frac{1-\mathrm{cos}2x}{2}dx$
$\frac{y}{\mathrm{sin}x}=\int \frac{1}{2}dx-\int \frac{\mathrm{cos}2x}{2}dx$
$\frac{y}{\mathrm{sin}x}=\frac{x}{2}-\frac{{\mathrm{sin}}^{2}n}{4}+c$
$y=\frac{2\mathrm{sin}x\mathrm{sin}x-\mathrm{sin}x\mathrm{sin}2x}{4}+c$
Jeffrey Jordon
Jeffrey Jordon
Jeffrey Jordon