Solve dy-(y-1)^2dx=0

$$d y - ( y - 1 ) ^ { 2 } d x = 0$$

Solve for d (complex solution)

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

Steps for Solving Linear Equation

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

$$dy-\left(y^{2}-2y+1\right)dx=0$$

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

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

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

$$dy-\left(y^{2}dx-2ydx+dx\right)=0$$

To find the opposite of y^{2}dx-2ydx+dx, find the opposite of each term.

$$dy-y^{2}dx+2ydx-dx=0$$

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$

Solve for x (complex solution)

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

Steps for Solving Linear Equation

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

$$dy-\left(y^{2}-2y+1\right)dx=0$$

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

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

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

$$dy-\left(y^{2}dx-2ydx+dx\right)=0$$

To find the opposite of y^{2}dx-2ydx+dx, find the opposite of each term.

$$dy-y^{2}dx+2ydx-dx=0$$

Subtract dy from both sides. Anything subtracted from zero gives its negation.

$$-y^{2}dx+2ydx-dx=-dy$$

Reorder the terms.

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

Combine all terms containing x.

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

Divide both sides by -dy^{2}+2dy-d.

$$\frac{\left(-dy^{2}+2dy-d\right)x}{-dy^{2}+2dy-d}=-\frac{dy}{-dy^{2}+2dy-d}$$

Dividing by -dy^{2}+2dy-d undoes the multiplication by -dy^{2}+2dy-d.

$$x=-\frac{dy}{-dy^{2}+2dy-d}$$

Divide -dy by -dy^{2}+2dy-d.

$$x=\frac{y}{\left(1-y\right)^{2}}$$

Solve for d

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

Steps for Solving Linear Equation

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

$$dy-\left(y^{2}-2y+1\right)dx=0$$

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

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

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

$$dy-\left(y^{2}dx-2ydx+dx\right)=0$$

To find the opposite of y^{2}dx-2ydx+dx, find the opposite of each term.

$$dy-y^{2}dx+2ydx-dx=0$$

Combine all terms containing d.

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

The equation is in standard form.

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

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

$$d=0$$

Solve for x

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

Steps for Solving Linear Equation

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

$$dy-\left(y^{2}-2y+1\right)dx=0$$

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

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

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

$$dy-\left(y^{2}dx-2ydx+dx\right)=0$$

To find the opposite of y^{2}dx-2ydx+dx, find the opposite of each term.

$$dy-y^{2}dx+2ydx-dx=0$$

Subtract dy from both sides. Anything subtracted from zero gives its negation.

$$-y^{2}dx+2ydx-dx=-dy$$

Reorder the terms.

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

Combine all terms containing x.

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

Divide both sides by -dy^{2}+2dy-d.

$$\frac{\left(-dy^{2}+2dy-d\right)x}{-dy^{2}+2dy-d}=-\frac{dy}{-dy^{2}+2dy-d}$$

Dividing by -dy^{2}+2dy-d undoes the multiplication by -dy^{2}+2dy-d.

$$x=-\frac{dy}{-dy^{2}+2dy-d}$$

Divide -dy by -dy^{2}+2dy-d.

$$x=\frac{y}{\left(1-y\right)^{2}}$$