# Initial Value Problem

• Jan 9th 2009, 06:40 AM
Rapha
Initial Value Problem
Hello, thanks for clicking my thread.

I need to solve the following IVP

$\displaystyle \dot{u} = (t+u)^2$ ;

u(0) = 0

Does anyone know how to solve this?

Kind regards
Rapha
• Jan 9th 2009, 07:45 AM
Mush
Quote:

Originally Posted by Rapha
Hello, thanks for clicking my thread.

I need to solve the following IVP

$\displaystyle \dot{u} = (t+u)^2$ ;

u(0) = 0

Does anyone know how to solve this?

Kind regards
Rapha

Ignore!
• Jan 9th 2009, 08:01 AM
Mush
Quote:

Originally Posted by Rapha
Hello, thanks for clicking my thread.

I need to solve the following IVP

$\displaystyle \dot{u} = (t+u)^2$ ;

u(0) = 0

Does anyone know how to solve this?

Kind regards
Rapha

Let $\displaystyle v = t + u$

Hence $\displaystyle u = v - t$

And $\displaystyle LHS = \frac{du}{dt} = \frac{d(v-t)}{dt} = \frac{dv}{dt} - \frac{dt}{dt} = \frac{dv}{dt} - 1$

And $\displaystyle RHS = v^2$

Hence $\displaystyle \frac{dv}{dt} - 1= v^2$

$\displaystyle \frac{dv}{dt} = v^2+1$

$\displaystyle \frac{dv}{v^2+1} = dt$

$\displaystyle \int \frac{dv}{v^2+1} = \int dt$

$\displaystyle \arctan{(v)} = t+C^*$

$\displaystyle v = \tan{(t+C^*)}$

Then using $\displaystyle \tan{(A+B)} = \frac{\tan{(A)}+\tan{(B)}}{1-\tan{(A)} \tan{(B)}}$, we find that:

$\displaystyle v = \frac{\tan{(t)}+\tan{(C^*)}}{1-\tan{(t)} \tan{(C^*)}} = \frac{\tan{(t)}+C}{1-C\tan{(A)}}$ where $\displaystyle C = \tan{(C^*)}$

And remember that $\displaystyle u = v-t$

$\displaystyle v-t= \frac{\tan{(t)}+C}{1-C\tan{(t)}} - t$

$\displaystyle u= \frac{\tan{(t)}+C - (t-Ct\tan{(t)})}{1-C\tan{(t)}}$

$\displaystyle u= \frac{\tan{(t)}+C - t+Ct\tan{(t)}}{1-C\tan{(t)}}$

$\displaystyle u= \frac{\tan{(t)}(1+Ct)+C - t}{1-C\tan{(t)}}$

Now apply your initial value conditions to find C!
• Jan 9th 2009, 08:20 AM
Mush
See corrections in above post!
• Jan 9th 2009, 07:10 PM
Rapha
Hi Mush.

Thank you so much, you are great!

Best wishes,
Rapha