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

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

Solve for d (complex solution)

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{C}\text{, }&x=\frac{\sqrt{1-8y^{2}}-1}{2}\text{ or }x=\frac{-\sqrt{1-8y^{2}}-1}{2}\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

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

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

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

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

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

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

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&\left(x=\frac{-\sqrt{1-8y^{2}}-1}{2}\text{ and }|y|\leq \frac{\sqrt{2}}{4}\right)\text{ or }\left(x=\frac{\sqrt{1-8y^{2}}-1}{2}\text{ and }|y|\leq \frac{\sqrt{2}}{4}\right)\text{ or }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

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

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

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

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

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

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

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$

Solve for x (complex solution)

$$\left\{\begin{matrix}\\x=\frac{-\sqrt{1-8y^{2}}-1}{2}\text{; }x=0\text{; }x=\frac{\sqrt{1-8y^{2}}-1}{2}\text{, }&\text{unconditionally}\\x\in \mathrm{C}\text{, }&d=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}\\x=0\text{, }&\text{unconditionally}\\x=\frac{\sqrt{1-8y^{2}}-1}{2}\text{; }x=\frac{-\sqrt{1-8y^{2}}-1}{2}\text{, }&|y|\leq \frac{\sqrt{2}}{4}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$