Question

Help!t%5E%7B2%7D%20%5Cfrac%7Bdy%7D%7Bdt%7D%2By%5E%7B2%7D%20%3Dty" id="TexFormula1" title="t^{2} \frac{dy}{dx}+y^{2} =ty\\t^{2} \frac{dy}{dt}+y^{2} =ty" alt="t^{2} \frac{dy}{dx}+y^{2} =ty\\t^{2} \frac{dy}{dt}+y^{2} =ty" align="absmiddle" class="latex-formula">

Answer 1

$t^2\dfrac{\mathrm dy}{\mathrm dt}+y^2=ty$

Divide both sides by $y^2$:

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

Let $z(t)=\dfrac t{y(t)}$, so that $y=\dfrac tz$ and

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{z-t\frac{\mathrm dz}{\mathrm dt}}{z^2}$

so that the ODE is transformed to

$z^2\dfrac{z-t\frac{\mathrm dz}{\mathrm dt}}{z^2}+1=z$

$z-t\dfrac{\mathrm dz}{\mathrm dt}+1=z$

and assuming $t>0$,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac1t$

The remaining ODE is separable:

$\mathrm dz=\dfrac{\mathrm dt}t\implies z=\ln t+C$

$\implies\dfrac ty=\ln t+C$

$\implies\dfrac yt=\dfrac1{\ln t+C}$

$\implies y=\dfrac t{\ln t+C}$