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-BARSIC- [3]
3 years ago
12

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

Mathematics
1 answer:
Sav [38]3 years ago
4 0

Replace y' with \dfrac{\mathrm dy}{\mathrm dx} to see that this ODE is separable:

2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\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=\dfrac{\mathrm dx}x. Then

\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\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\sqrt{C|\ln x|-4}

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