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

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

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply y and y to get y^{2}.

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

Use the distributive property to multiply x^{2}+y^{2} by d.

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

Use the distributive property to multiply x^{2}d+y^{2}d by x.

$$dx^{3}+y^{2}dx+3xy^{2}d=0$$

Combine y^{2}dx and 3xy^{2}d to get 4y^{2}dx.

$$dx^{3}+4y^{2}dx=0$$

Combine all terms containing d.

$$\left(x^{3}+4y^{2}x\right)d=0$$

The equation is in standard form.

$$\left(x^{3}+4xy^{2}\right)d=0$$

Divide 0 by x^{3}+4y^{2}x.

$$d=0$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$