Solve the differential equation dy = dx/(y - 1)^2.

Step-by-step solution

1. 1

   Rewrite it in derivative form. Dividing through by $dx$:

   $\frac{dy}{dx}=\frac{1}{(y-1)^2}$

   The right side depends on $y$ alone, so the equation is separable. Note the restriction $y\neq1$, where the right side is undefined.

2. 2

   Separate the variables. Multiply both sides by $(y-1)^2$ and by $dx$:

   $(y-1)^2\,dy=dx$

   The squared factor moves to the $y$ side, which is what makes the integral elementary.

3. 3

   Integrate both sides. Substituting $u=y-1$ (so $du=dy$) turns the left integral into a plain power:

   $\int(y-1)^2dy=\int u^2du=\frac{u^3}{3}=\frac{(y-1)^3}{3},\qquad \int dx=x+C$

   so

   $\frac{(y-1)^3}{3}=x+C$

4. 4

   Clear the fraction. Multiplying by $3$ absorbs the constant ($3C$ is just as arbitrary as $C$):

   $(y-1)^3=3x+C$

   This is the implicit general solution.

5. 5

   Solve for y explicitly. Cube roots are single-valued over the reals — unlike square roots, there is no $\pm$ and no sign case to consider:

   $y-1=\sqrt[3]{3x+C}\qquad\Longrightarrow\qquad y=1+\sqrt[3]{3x+C}$

   The solution passes through $y=1$ exactly where $3x+C=0$; there the derivative is infinite, so the curve has a vertical tangent.

6. 6

   Verify by differentiating implicitly. From $(y-1)^3=3x+C$: $3(y-1)^2y'=3$, so $y'=\dfrac{1}{(y-1)^2}$ ✓ — the original equation. Numerically with $C=2$ at $x=0.5,\,2.0,\,4.0$, the difference quotient of $y=1+\sqrt[3]{3x+2}$ matches $\tfrac{1}{(y-1)^2}$ to four digits ✓.