Solve the differential equation dy/dx - y^2 = 0.

Step-by-step solution

1. 1

   Rearrange into standard separable form. Moving the $y^2$ across gives

   $\frac{dy}{dx}=y^2$

   The right side depends on $y$ only, so the equation separates. Note also the constant solution $y=0$, which satisfies the equation trivially and will have to be treated apart.

2. 2

   Separate the variables. Assuming $y\neq0$, divide by $y^2$:

   $\frac{dy}{y^2}=dx,\qquad\text{i.e.}\qquad y^{-2}\,dy=dx$

3. 3

   Integrate both sides with the power rule. For $n=-2$, $\int y^{n}dy=\frac{y^{n+1}}{n+1}$ gives $\frac{y^{-1}}{-1}=-y^{-1}$ — the $\ln$ case does not apply here because the exponent is not $-1$:

   $-\frac{1}{y}=x+C$

4. 4

   Solve for y. Multiply by $-1$, then rename the constant ($-C$ is just as arbitrary as $C$):

   $\frac{1}{y}=-x-C=C'-x$

   Taking reciprocals,

   $y=\frac{1}{C-x}$

   writing $C$ for $C'$. Unlike the linear case $y'=y$, here $C=0$ does not recover $y=0$; the zero solution is a genuinely separate singular solution lost in step 2.

5. 5

   Note the finite-time blow-up. The solution has a vertical asymptote at $x=C$: as $x\to C^-$ the value $|y|\to\infty$. So no non-zero solution exists on the whole real line — a hallmark of quadratic growth that linear equations never show.

6. 6

   Verify by differentiation. For $y=\frac{1}{C-x}=(C-x)^{-1}$, the chain rule gives $y'=(C-x)^{-2}=y^2$ ✓. Numerically with $C=2$ at $x=0,\,0.5,\,-1.3$, the symmetric difference quotient matches $y^2$ to five digits ✓.