Question

Find a general solution of
t *(dy/dt)-(y^2)*lnt+y=0

Answer 1

`t(\mathrm dy)/(\mathrm dt)-y^2\ln t+y=0`

Divide both sides by
`y(t)^2`:

`ty^(-2)(\mathrm dy)/(\mathrm dt)-\ln t+y^(-1)=0`

Substitute
`v(t)=y(t)^(-1)`, so that
`(\mathrm dv)/(\mathrm dt)=-y(t)^(-2)(\mathrm dy)/(\mathrm dt)`.

`-t(\mathrm dv)/(\mathrm dt)-\ln t+v=0`

`t(\mathrm dv)/(\mathrm dt)-v=\ln t`

Divide both sides by
`t^2`:

`\frac1t(\mathrm dv)/(\mathrm dt)-\frac1{t^2}v=(\ln t)/(t^2)`

The left side can be condensed as the derivative of a product:

`(\mathrm d)/(\mathrm dt)\left[\frac1tv\right]=(\ln t)/(t^2)`

Integrate both sides. The integral on the right side can be done by parts.

`\displaystyle\int(\ln t)/(t^2)\,\mathrm dt=-\frac{\ln t}t+\int(\mathrm dt)/(t^2)=-\frac{\ln t}t-\frac1t+C`

`\frac1tv=-\frac{\ln t}t-\frac1t+C`

`v=-\ln t-1+Ct`

Now solve for
`y(t)`.

`y^(-1)=-\ln t-1+Ct`

`\boxed{y(t)=\frac1{Ct-\ln t-1}}`