Question

How do you solve this differential equation t^2dy/dx+y^2=ty

Answer 1

If
`t` is just a constant, then you have a fairly simple (separable) ODE.


`t^2(\mathrm dy)/(\mathrm dx)+y^2=ty\implies (\mathrm dy)/(ty-y^2)=(\mathrm dx)/(t^2)`

and so on.

So, I'll assume you meant to write
`(\mathrm dy)/(\mathrm dt)`...


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

This is a standard Bernoulli equation, which means a substitution of
`y=z^(1-2)=z^(-1)` will suffice to transform this ODE in
`y` into a linear ODE in
`z`. You have


`(\mathrm dy)/(\mathrm dt)=(\mathrm dy)/(\mathrm dz)*(\mathrm dz)/(\mathrm dt)`

`(\mathrm dy)/(\mathrm dt)=-\frac1{z^2}(\mathrm dz)/(\mathrm dt)`

which changes the ODE to


`-(t^2)/(z^2)(\mathrm dz)/(\mathrm dt)+\frac1{z^2}=\frac tz`

`(\mathrm dz)/(\mathrm dt)+\frac1tz=\frac1{t^2}`

An integrating factor would be


`\mu(t)=\exp\left(\displaystyle\int\frac{\mathrm dt}t\right)=t`

Multiplying both sides by the IF gives


`t(\mathrm dz)/(\mathrm dt)+z=\frac1t`

`(\mathrm d)/(\mathrm dt)[tz]=\frac1t`

Integrate both sides wrt
`t` to get


`tz=\displaystyle\int\frac{\mathrm dt}t=\ln|t|`

`z=\fractt`

Now back substitute. Since
`y=\frac1z`, you get
`z=\frac1y` and the solution is


`y=\frac tt`