Solve (e^2ycosx-ycosx)dy-(e^ysin2x)dx=0

$$( e ^ { 2 y } \cos x - y \cos x ) d y - ( e ^ { y } \sin 2 x ) d x = 0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&y\left(\left(e^{y}\right)^{2}-y\right)-2x\sin(x)e^{y}=0\text{ or }\exists n_{1}\in \mathrm{Z}\text{ : }x=\frac{\pi \left(2n_{1}+1\right)}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply e^{2y}\cos(x)-y\cos(x) by d.

$$\left(e^{2y}\cos(x)d-y\cos(x)d\right)y-e^{y}\sin(2x)dx=0$$

Use the distributive property to multiply e^{2y}\cos(x)d-y\cos(x)d by y.

$$e^{2y}\cos(x)dy-\cos(x)dy^{2}-e^{y}\sin(2x)dx=0$$

Combine all terms containing d.

$$\left(e^{2y}\cos(x)y-\cos(x)y^{2}-e^{y}\sin(2x)x\right)d=0$$

The equation is in standard form.

$$\left(y\cos(x)e^{2y}-x\sin(2x)e^{y}-y^{2}\cos(x)\right)d=0$$

Divide 0 by e^{2y}\cos(x)y-\cos(x)y^{2}-e^{y}\sin(2x)x.

$$d=0$$