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

b. The ODE is separable as

`(y^2+1)(\mathrm dy)/(\mathrm dx)=3yx^2\implies\frac{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`