Solve (cosxcosy+2x)dx-(sinxsiny+2y)dy=0

$$( \cos x \cos y + 2 x ) d x - ( \sin x \sin y + 2 y ) d y = 0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x\left(\cos(x)\cos(y)+2x\right)-y\left(\sin(x)\sin(y)+2y\right)=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply \cos(x)\cos(y)+2x by d.

$$\left(\cos(x)\cos(y)d+2xd\right)x-\left(\sin(x)\sin(y)+2y\right)dy=0$$

Use the distributive property to multiply \cos(x)\cos(y)d+2xd by x.

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

Use the distributive property to multiply \sin(x)\sin(y)+2y by d.

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

Use the distributive property to multiply \sin(x)\sin(y)d+2yd by y.

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

To find the opposite of \sin(x)\sin(y)dy+2dy^{2}, find the opposite of each term.

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

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$