Question

The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Answer 1

Replace
`y'` with
`(\mathrm dy)/(\mathrm dx)` to see that this ODE is separable:

`2y\ln x(\mathrm dy)/(\mathrm dx)=\frac{y^2+4}x\implies(2y)/(y^2+4)\,\mathrm dy=(\mathrm dx)/(x\ln x)`

Integrate both sides; on the left, set
`u=y^2+4` so that
`\mathrm du=2y\,\mathrm dy`; on the right, set
`v=\ln x` so that
`\mathrm dv=\frac{\mathrm dx}x`. Then

`\displaystyle\int(2y)/(y^2+4)\,\mathrm dy=\int(\mathrm dx)/(x\ln x)\iff\int\frac{\mathrm du}u=\int\frac{\mathrm dv}v`

`\implies\ln|u|=\ln|v|+C`

`\implies\ln(y^2+4)=\ln|\ln x|+C`

`\implies y^2+4=e^(\ln|\ln x|+C)`

`\implies y^2=C|\ln x|-4`

`\implies y=\pm√(C|\ln x|-4)`