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

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

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=-\sqrt[3]{1-3y}y^{\frac{2}{3}}\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

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

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

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

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

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$

Solve for x

$$\left\{\begin{matrix}\\x=-\sqrt[3]{1-3y}y^{\frac{2}{3}}\text{; }x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$