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Question

Help!

Answer 1

`t^2(\mathrm dy)/(\mathrm dt)+y^2=ty`

Divide both sides by
`y^2`:

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

Let
`z(t)=\frac t{y(t)}`, so that
`y=\frac tz` and

`(\mathrm dy)/(\mathrm dt)=(z-t(\mathrm dz)/(\mathrm dt))/(z^2)`

so that the ODE is transformed to

`z^2(z-t(\mathrm dz)/(\mathrm dt))/(z^2)+1=z`

`z-t(\mathrm dz)/(\mathrm dt)+1=z`

and assuming
`t>0`,

`(\mathrm dz)/(\mathrm dt)=\frac1t`

The remaining ODE is separable:

`\mathrm dz=\frac{\mathrm dt}t\implies z=\ln t+C`

`\implies\frac ty=\ln t+C`

`\implies\frac yt=\frac1{\ln t+C}`

`\implies y=\frac t{\ln t+C}`