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

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

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x=\sqrt[3]{\frac{\sqrt{69}\left(|y|\right)^{3}}{18}-\frac{25y^{3}}{54}}+\sqrt[3]{-\frac{\sqrt{69}\left(|y|\right)^{3}}{18}-\frac{25y^{3}}{54}}+\frac{y}{3}\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

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

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

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

$$dx^{4}+y^{3}dx-x^{3}yd=0$$

Reorder the terms.

$$dx^{4}+dxy^{3}-dyx^{3}=0$$

Combine all terms containing d.

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

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

$$d=0$$

Solve for x

$$\left\{\begin{matrix}\\x=\frac{2^{\frac{2}{3}}\left(\sqrt[3]{3\sqrt{69}\left(|y|\right)^{3}-25y^{3}}+\sqrt[3]{-3\sqrt{69}\left(|y|\right)^{3}-25y^{3}}+\sqrt[3]{2}y\right)}{6}\text{; }x=0\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$