Question

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

Answer 1

$t\dfrac{\mathrm dy}{\mathrm dt}-y^2\ln t+y=0$

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

$ty^{-2}\dfrac{\mathrm dy}{\mathrm dt}-\ln t+y^{-1}=0$

Substitute $v(t)=y(t)^{-1}$, so that $\dfrac{\mathrm dv}{\mathrm dt}=-y(t)^{-2}\dfrac{\mathrm dy}{\mathrm dt}$.

$-t\dfrac{\mathrm dv}{\mathrm dt}-\ln t+v=0$

$t\dfrac{\mathrm dv}{\mathrm dt}-v=\ln t$

Divide both sides by $t^2$:

$\dfrac1t\dfrac{\mathrm dv}{\mathrm dt}-\dfrac1{t^2}v=\dfrac{\ln t}{t^2}$

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

$\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac1tv\right]=\dfrac{\ln t}{t^2}$

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

$\displaystyle\int\frac{\ln t}{t^2}\,\mathrm dt=-\frac{\ln t}t+\int\frac{\mathrm dt}{t^2}=-\frac{\ln t}t-\frac1t+C$

$\dfrac1tv=-\dfrac{\ln t}t-\dfrac1t+C$

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

Now solve for $y(t)$.

$y^{-1}=-\ln t-1+Ct$

$\boxed{y(t)=\dfrac1{Ct-\ln t-1}}$