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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+2x^{2}\right)^{2}.

$$\left(y^{2}+4yx^{2}+4\left(x^{2}\right)^{2}\right)dx-x^{3}dy=0$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\left(y^{2}+4yx^{2}+4x^{4}\right)dx-x^{3}dy=0$$

Use the distributive property to multiply y^{2}+4yx^{2}+4x^{4} by d.

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

Use the distributive property to multiply y^{2}d+4yx^{2}d+4x^{4}d by x.

$$y^{2}dx+4ydx^{3}+4dx^{5}-x^{3}dy=0$$

Combine 4ydx^{3} and -x^{3}dy to get 3ydx^{3}.

$$y^{2}dx+3ydx^{3}+4dx^{5}=0$$

Combine all terms containing d.

$$\left(y^{2}x+3yx^{3}+4x^{5}\right)d=0$$

The equation is in standard form.

$$\left(4x^{5}+xy^{2}+3yx^{3}\right)d=0$$

Divide 0 by y^{2}x+3yx^{3}+4x^{5}.

$$d=0$$

Solve for x

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