Question

Consider the ODE

(y^2 + 1)y' = 3yx^2

a) Find the equilibrium solutions

b) Solve the initial value problem if y(0) = 1

Answer 1

a. The equilibrium solutions occur where $y'=0$:

$(y^2+1)y'=3yx^2\implies y'=\dfrac{3yx^2}{y^2+1}=0\implies x=0\text{ or }y=0$

b. The ODE is separable as

$(y^2+1)\dfrac{\mathrm dy}{\mathrm dx}=3yx^2\implies\dfrac{y^2+1}y\,\mathrm dy=3x^2\,\mathrm dx$

Integrate both sides to get

$\displaystyle\int\frac{y^2+1}y\,\mathrm dy=3\int x^2\,\mathrm dx$

$\displaystyle\int\left(y+\frac1y\right)\,\mathrm dy=3\int x^2\,\mathrm dx$

$\displaystyle\frac{y^2}2+\ln|y|=x^3+C$