Initial Value Problem

**Rapha** · January 9th 2009, 06:40 AM

Hello, thanks for clicking my thread.

I need to solve the following IVP

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

u(0) = 0

Does anyone know how to solve this?

Kind regards
Rapha

**Mush** · January 9th 2009, 07:45 AM

Originally Posted by Rapha

Hello, thanks for clicking my thread.

I need to solve the following IVP

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

u(0) = 0

Does anyone know how to solve this?

Kind regards
Rapha

Ignore!

**Mush** · January 9th 2009, 08:01 AM

Originally Posted by Rapha

Hello, thanks for clicking my thread.

I need to solve the following IVP

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

u(0) = 0

Does anyone know how to solve this?

Kind regards
Rapha

Let $v = t + u$

Hence $u = v - t$

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

And $RHS = v^2$

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

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

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

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

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

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

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

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

And remember that $u = v-t$

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

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

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

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

Now apply your initial value conditions to find C!

**Mush** · January 9th 2009, 08:20 AM

See corrections in above post!

**Rapha** · January 9th 2009, 07:10 PM

Hi Mush.

Thank you so much, you are great!

Best wishes,
Rapha

Thread: Initial Value Problem

LinkBack

Thread Tools

Display

Initial Value Problem

Similar Math Help Forum Discussions

DE Initial Value Problem

initial value problem

Initial value problem

Initial Value Problem?

Initial Value Problem

Search Tags