# {y}'- \sin{{2}}{x}+{y} \tan{{x}}={0}

${y}^{\prime }-\mathrm{sin}2x+y\mathrm{tan}x=0$

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Solve the linear equation $\frac{dy\left(x\right)}{dx}-\mathrm{sin}\left(2x\right)+\mathrm{tan}\left(x\right)y\left(x\right)=0$:
Add $\mathrm{sin}\left(2x\right)$ to both sides:
$\frac{dy\left(x\right)}{dx}+\mathrm{tan}\left(x\right)y\left(x\right)=\mathrm{sin}\left(2x\right)$
Let $\mu \left(x\right)={e}^{\int \mathrm{tan}\left(x\right)dx}=\mathrm{sec}\left(x\right)$.
Multiply both sides by $\mu \left(x\right)$:
$\mathrm{sec}\left(x\right)\frac{dy\left(x\right)}{dx}+\left(\mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)\right)y\left(x\right)=\mathrm{sec}\left(x\right)\mathrm{sin}\left(2x\right)$
INTERMEDIATE STEPS:
Take the integral:
$\int \mathrm{tan}\left(x\right)dx$
Rewrite :
$\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$
For the integrand $\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}$, substitute $u=\mathrm{cos}\left(x\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}du=-\mathrm{sin}\left(x\right)dx$:
$=\int -\frac{1}{u}du$
Factor out constants:
$=\int -\frac{1}{u}du$
The integral of $\frac{1}{u}$ is $\mathrm{log}\left(u\right)$:
$=-\mathrm{log}\left(u\right)+\text{constant}$
Substitute back for $u=\mathrm{cos}\left(x\right)$:
$=-\mathrm{log}\left(\mathrm{cos}\left(x\right)\right)+\text{constant}$
Substitute $\mathrm{tan}\left(x\right)\mathrm{sec}\left(x\right)=\frac{d\mathrm{sec}\left(x\right)}{dx}$:
$\mathrm{sec}\left(x\right)\frac{dy\left(x\right)}{dx}+\frac{d\mathrm{sec}\left(x\right)}{dx}y\left(x\right)=\mathrm{sec}\left(x\right)\mathrm{sin}\left(2x\right)$
INTERMEDIATE STEPS:
Possible derivation:
$\frac{d}{dx}\left(\mathrm{sec}\left(x\right)\right)$
Rewrite the expression: