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expeople1 [14]
3 years ago
8

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

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0

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}}

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