## Find solution of initial value problem - 1st order non-linear ODE

Hey,
we have to solve the following problem for our ODE class.

1. The problem statement, all variables and given/known data

Find the solution of the initial value problem
dx/dt = (x^2 + t*x - t^2)/t^2

with t≠0 , x(t_0) = x_0

Describe the (maximal) domain of definition of the solution.

3. The attempt at a solution
Well, I know that this is a 1st order nonlinear ODE. Unfortunately I got no clue how to deal them.
I tried this:
dx/dt = (x^2 + t*x - t^2)/t^2
= x^2/t^2 + x/t -1

Now substitute: u = x/t -> x=ut , x'=u't+u
Therefore we get:
u't+u = u^2+u-1
t* du/dt +u = u^2+u-1 //-u
t* du/dt = u^2 -1

0= t*u' -u^2 +1

Is the idea ok? What could I do?

Kind regards,
mihyaeru

PS: How can i insert a fraction?

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 I just noticed that we might be able to solve this via applying the Riccati equation????

Recognitions:
Homework Help
 Quote by mihyaeru t* du/dt = u^2 -1
This is a separable differential equation, you must be able to solve it
!

ehild

## Find solution of initial value problem - 1st order non-linear ODE

As far as I know it is a separation of variables case, iff there would be no t in front of the du/dt. Or no -1.
You think of the solution
du / (u^2 -1) = dt / t ,
don't you?

Recognitions:
Gold Member
Homework Help
 Quote by mihyaeru As far as I know it is a separation of variables case, iff there would be no t in front of the du/dt.
That doesn't matter. Just keep the t with the dt.
 You think of the solution du / (u^2 -1) = dt / t , don't you?
That's exactly what he is thinking of. Use partial fractions on the left side and integrate both sides.

Recognitions:
Homework Help
 Quote by mihyaeru As far as I know it is a separation of variables case, iff there would be no t in front of the du/dt. Or no -1. You think of the solution du / (u^2 -1) = dt / t , don't you? But anyway thanks for your answer =)
I do. And you should be able to integrate both sides.

ehild

 Tags 2nd semester, math courses, ode, ode first order help, ode non linear