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

 P: 4 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 which is my dead end. Is the idea ok? What could I do? Kind regards, mihyaeru PS: How can i insert a fraction?
 P: 4 I just noticed that we might be able to solve this via applying the Riccati equation????
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P: 10,662
 Quote by mihyaeru t* du/dt = u^2 -1
This is a separable differential equation, you must be able to solve it
!

ehild

 P: 4 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? But anyway thanks for your answer =)
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P: 7,662
 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.
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P: 10,662
 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

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