[tex]t^2\dfrac{\mathrm dy}{\mathrm dt}+y^2=ty[/tex]
Divide both sides by [tex]y^2[/tex]:
[tex]\dfrac{t^2}{y^2}\dfrac{\mathrm dy}{\mathrm dt}+1=\dfrac ty[/tex]
Let [tex]z(t)=\dfrac t{y(t)}[/tex], so that [tex]y=\dfrac tz[/tex] and
[tex]\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{z-t\frac{\mathrm dz}{\mathrm dt}}{z^2}[/tex]
so that the ODE is transformed to
[tex]z^2\dfrac{z-t\frac{\mathrm dz}{\mathrm dt}}{z^2}+1=z[/tex]
[tex]z-t\dfrac{\mathrm dz}{\mathrm dt}+1=z[/tex]
and assuming [tex]t>0[/tex],
[tex]\dfrac{\mathrm dz}{\mathrm dt}=\dfrac1t[/tex]
The remaining ODE is separable:
[tex]\mathrm dz=\dfrac{\mathrm dt}t\implies z=\ln t+C[/tex]
[tex]\implies\dfrac ty=\ln t+C[/tex]
[tex]\implies\dfrac yt=\dfrac1{\ln t+C}[/tex]
[tex]\implies y=\dfrac t{\ln t+C}[/tex]
