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

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

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