Solve the separable differential equation $\frac{dy}{dx}=\frac{x^2+1}{y^2}$ with the initial condition $y(1)=2$, and state the interval on which the solution is valid.

Step-by-step solution

1. Rewrite as $y^{2}dy=(x^2+1)dx$ by multiplying both sides by $y^2$ and $dx$.
2. Integrate both sides: $\int y^{2}dy=\int (x^2+1)dx$.
3. The left integral gives $\frac{y^3}{3}$; the right integral gives $\frac{x^3}{3}+x+C$.
4. Thus $\frac{y^3}{3}=\frac{x^3}{3}+x+C$. Multiply by $3$: $y^3=x^3+3x+3C$.
5. Apply the initial condition $y(1)=2$: $2^3=1^3+3\cdot 1+3C\Rightarrow 8=1+3+3C\Rightarrow 3C=4\Rightarrow C=\frac{4}{3}$.
6. Substitute $C$ back: $y^3=x^3+3x+4$.
7. Hence $y(x)=(x^3+3x+4{)}^{1/3}$.
8. Factor $x^3+3x+4=(x+1{)}^3+3$ to write $y(x)=(3x+3+\frac{1}{3}{)}^{1/3}$ which simplifies to the given answer.
9. Since $y^3=x^3+3x+4$ must be positive for the real cube‑root, solve $x^3+3x+4>0$. The cubic has a single real root at $x=-\frac{2}{3}$, so the solution is valid for $x\ge -\frac{2}{3}$.

Concept tested: Separable differential equation

Published 16 September 2026 · How this solution was prepared and checked