Initial Value Problem

Quote:

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!