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

  • Thread starter mihyaeru
  • Start date
  • #1
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Hey,
we have to solve the following problem for our ODE class.

Homework Statement



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.

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?
 

Answers and Replies

  • #2
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I just noticed that we might be able to solve this via applying the Riccati equation????
 
  • #3
ehild
Homework Helper
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t* du/dt = u^2 -1
This is a separable differential equation, you must be able to solve it
:smile: !

ehild
 
  • #4
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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 =)
 
  • #5
LCKurtz
Science Advisor
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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.
 
  • #6
ehild
Homework Helper
15,543
1,909
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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