Solve 2y(x+1)dy=xdx

$$2 y ( x + 1 ) d y = x d x$$

Solve for d (complex solution)

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

Steps for Solving Linear Equation

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

$$2y^{2}\left(x+1\right)d=xdx$$

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

$$2y^{2}\left(x+1\right)d=x^{2}d$$

Use the distributive property to multiply 2y^{2} by x+1.

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

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

$$2y^{2}xd+2y^{2}d=x^{2}d$$

Subtract x^{2}d from both sides.

$$2y^{2}xd+2y^{2}d-x^{2}d=0$$

Reorder the terms.

$$-dx^{2}+2dxy^{2}+2dy^{2}=0$$

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&x>-1\text{ and }|y|=\frac{|x|}{\sqrt{2x+2}}\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2y^{2}\left(x+1\right)d=xdx$$

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

$$2y^{2}\left(x+1\right)d=x^{2}d$$

Use the distributive property to multiply 2y^{2} by x+1.

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

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

$$2y^{2}xd+2y^{2}d=x^{2}d$$

Subtract x^{2}d from both sides.

$$2y^{2}xd+2y^{2}d-x^{2}d=0$$

Reorder the terms.

$$-dx^{2}+2dxy^{2}+2dy^{2}=0$$

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$

Solve for x (complex solution)

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

Solve for x

$$\left\{\begin{matrix}\\x=y\left(-\sqrt{y^{2}+2}+y\right)\text{; }x=y\left(\sqrt{y^{2}+2}+y\right)\text{, }&\text{unconditionally}\\x\in \mathrm{R}\text{, }&d=0\end{matrix}\right.$$