Solve cos(x+y)dx+(3y^2+2y+cos(x+y))dy=0

$$\cos ( x + y ) d x + ( 3 y ^ { 2 } + 2 y + \cos ( x + y ) ) d y = 0$$

Solve for d (complex solution)

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{C}\text{, }&y\left(\frac{1}{e^{ix+iy}}+e^{ix+iy}+6y^{2}+4y\right)+x\left(\frac{1}{e^{ix+iy}}+e^{ix+iy}\right)=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3y^{2}+2y+\cos(x+y) by d.

$$\cos(x+y)dx+\left(3y^{2}d+2yd+\cos(x+y)d\right)y=0$$

Use the distributive property to multiply 3y^{2}d+2yd+\cos(x+y)d by y.

$$\cos(x+y)dx+3dy^{3}+2dy^{2}+\cos(x+y)dy=0$$

Combine all terms containing d.

$$\left(\cos(x+y)x+3y^{3}+2y^{2}+\cos(x+y)y\right)d=0$$

The equation is in standard form.

$$\left(x\cos(x+y)+y\cos(x+y)+3y^{3}+2y^{2}\right)d=0$$

Divide 0 by \cos(x+y)x+3y^{3}+2y^{2}+\cos(x+y)y.

$$d=0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&\left(x+y\right)\cos(x+y)+3y^{3}+2y^{2}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use the distributive property to multiply 3y^{2}+2y+\cos(x+y) by d.

$$\cos(x+y)dx+\left(3y^{2}d+2yd+\cos(x+y)d\right)y=0$$

Use the distributive property to multiply 3y^{2}d+2yd+\cos(x+y)d by y.

$$\cos(x+y)dx+3dy^{3}+2dy^{2}+\cos(x+y)dy=0$$

Combine all terms containing d.

$$\left(\cos(x+y)x+3y^{3}+2y^{2}+\cos(x+y)y\right)d=0$$

The equation is in standard form.

$$\left(x\cos(x+y)+y\cos(x+y)+3y^{3}+2y^{2}\right)d=0$$

Divide 0 by \cos(x+y)x+3y^{3}+2y^{2}+\cos(x+y)y.

$$d=0$$