Solve dy=(x+1)^2-1

$$d y = ( x + 1 ) ^ { 2 } - 1$$

Solve for d (complex solution)

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

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$dy=x^{2}+2x+1-1$$

Subtract 1 from 1 to get 0.

$$dy=x^{2}+2x$$

The equation is in standard form.

$$yd=x^{2}+2x$$

Divide both sides by y.

$$\frac{yd}{y}=\frac{x\left(x+2\right)}{y}$$

Dividing by y undoes the multiplication by y.

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

Solve for d

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

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$dy=x^{2}+2x+1-1$$

Subtract 1 from 1 to get 0.

$$dy=x^{2}+2x$$

The equation is in standard form.

$$yd=x^{2}+2x$$

Divide both sides by y.

$$\frac{yd}{y}=\frac{x\left(x+2\right)}{y}$$

Dividing by y undoes the multiplication by y.

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

Solve for x (complex solution)

$$x=-\left(\sqrt{dy+1}+1\right)$$

$$x=\sqrt{dy+1}-1$$

Solve for x

$$\left\{\begin{matrix}x=-1\text{, }&d=-\frac{1}{y}\text{ and }y\neq 0\\x=-\left(\sqrt{dy+1}+1\right)\text{; }x=\sqrt{dy+1}-1\text{, }&\left(d\geq -\frac{1}{y}\text{ and }y>0\right)\text{ or }\left(d\leq -\frac{1}{y}\text{ and }y<0\right)\\x=-2\text{; }x=0\text{, }&y=0\end{matrix}\right.$$