Solve xpartialx+(y-2x)dy=0

$$x \partial x + ( y - 2 x ) d y = 0$$

Solve for d (complex solution)

$$\left\{\begin{matrix}d=-\frac{∂x^{2}}{y\left(y-2x\right)}\text{, }&y\neq 0\text{ and }x\neq \frac{y}{2}\\d\in \mathrm{C}\text{, }&\left(x=0\text{ or }∂=0\right)\text{ and }\left(y=0\text{ or }x=\frac{y}{2}\right)\text{ and }\left(y=0\text{ or }∂=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}∂+\left(y-2x\right)dy=0$$

Use the distributive property to multiply y-2x by d.

$$x^{2}∂+\left(yd-2xd\right)y=0$$

Use the distributive property to multiply yd-2xd by y.

$$x^{2}∂+dy^{2}-2xdy=0$$

Subtract x^{2}∂ from both sides. Anything subtracted from zero gives its negation.

$$dy^{2}-2xdy=-x^{2}∂$$

Reorder the terms.

$$-2dxy+dy^{2}=-∂x^{2}$$

Combine all terms containing d.

$$\left(-2xy+y^{2}\right)d=-∂x^{2}$$

The equation is in standard form.

$$\left(y^{2}-2xy\right)d=-∂x^{2}$$

Divide both sides by y^{2}-2yx.

$$\frac{\left(y^{2}-2xy\right)d}{y^{2}-2xy}=-\frac{∂x^{2}}{y^{2}-2xy}$$

Dividing by y^{2}-2yx undoes the multiplication by y^{2}-2yx.

$$d=-\frac{∂x^{2}}{y^{2}-2xy}$$

Divide -∂x^{2} by y^{2}-2yx.

$$d=-\frac{∂x^{2}}{y\left(y-2x\right)}$$

Solve for d

$$\left\{\begin{matrix}d=-\frac{∂x^{2}}{y\left(y-2x\right)}\text{, }&y\neq 0\text{ and }x\neq \frac{y}{2}\\d\in \mathrm{R}\text{, }&\left(x=0\text{ or }∂=0\right)\text{ and }\left(y=0\text{ or }x=\frac{y}{2}\right)\text{ and }\left(y=0\text{ or }∂=0\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$x^{2}∂+\left(y-2x\right)dy=0$$

Use the distributive property to multiply y-2x by d.

$$x^{2}∂+\left(yd-2xd\right)y=0$$

Use the distributive property to multiply yd-2xd by y.

$$x^{2}∂+dy^{2}-2xdy=0$$

Subtract x^{2}∂ from both sides. Anything subtracted from zero gives its negation.

$$dy^{2}-2xdy=-x^{2}∂$$

Reorder the terms.

$$-2dxy+dy^{2}=-∂x^{2}$$

Combine all terms containing d.

$$\left(-2xy+y^{2}\right)d=-∂x^{2}$$

The equation is in standard form.

$$\left(y^{2}-2xy\right)d=-∂x^{2}$$

Divide both sides by y^{2}-2yx.

$$\frac{\left(y^{2}-2xy\right)d}{y^{2}-2xy}=-\frac{∂x^{2}}{y^{2}-2xy}$$

Dividing by y^{2}-2yx undoes the multiplication by y^{2}-2yx.

$$d=-\frac{∂x^{2}}{y^{2}-2xy}$$

Divide -∂x^{2} by y^{2}-2yx.

$$d=-\frac{∂x^{2}}{y\left(y-2x\right)}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{dy+\sqrt{d\left(d-∂\right)y^{2}}}{∂}\text{; }x=\frac{dy-\sqrt{d\left(d-∂\right)y^{2}}}{∂}\text{, }&∂\neq 0\\x=\frac{y}{2}\text{, }&∂=0\text{ and }y\neq 0\text{ and }d\neq 0\\x\in \mathrm{C}\text{, }&\left(d=0\text{ or }y=0\right)\text{ and }∂=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{y\left(\sqrt{d\left(d-∂\right)}+d\right)}{∂}\text{; }x=\frac{y\left(-\sqrt{d\left(d-∂\right)}+d\right)}{∂}\text{, }&∂\neq 0\text{ and }\left(d\geq 0\text{ or }∂\geq d\right)\text{ and }\left(d\leq 0\text{ or }∂\leq d\right)\\x=\frac{y}{2}\text{, }&y=0\text{ or }\left(∂=0\text{ and }y\neq 0\text{ and }d\neq 0\right)\\x\in \mathrm{R}\text{, }&\left(d=0\text{ or }y=0\right)\text{ and }∂=0\end{matrix}\right.$$