Homework for holidays -2

[Camillo]

Solve the Cauchy problem :

$u''(x)+2u'(x)+2u(x) = f(x) $ , for $ x> 0$
$u(0)=u'(0) =0 $ .

[Paolo902]

"Kroldar":I think that's right.

Very good news (I am so happy!). Thanks, Kroldar, for your post and for everything you have taught me (indeed, your lessons about Laplace transform have been very useful).
Thanks,

[Kroldar]

I think that's right.

[Paolo902]

Good to hear from you again, Camillo.

Thank you for giving us this kind of homework, I do love it. Unfortunately this week I have been very busy and therefore I haven't been able to do it. I'll try to do it now, so let's see.

Have to say that when I first saw the equation I thought it was "impossible". I thought that we should have known what kind of function was $f(x)$. Indeed, I solved the differential homogeneous equation ($u''+2u'+2u=0$) but I didn't find it very useful.

Then I had the "great" idea (:D): what about using Laplace transform? I think this should be correct, also because you have given us the initial conditions. I use the spoiler (so everybody can try to find the solution):

$u''+2u'+2u=f(x)$, $x>0$

Applying Laplace (unilateral, since $x>0$) transform to both sides we get
$ccL(u'')+2ccL(u')+2ccL(u)=ccL[f(x)](s)$

Now, let us call $ccL[u(x)](s)=U(s)$; then, remembering that $ccL[f'(x)](s)= s*ccL(f)(s)-f(0)$ and that $ccL[f''(x)](s)= s^2*ccL(f)(s)-sf(0)-f'(0)$ and that $u(0)=u'(0)=0$ our equation becomes
$s^2U+2sU+2U=ccL[f(x)](s)$

By algebraic manipulation we obtain
$U(s)=\frac{ccL[f(x)](s)}{s^2+2s+2}$

We can write it in another way:
$U(s)=ccL[f(x)](s)*\frac{1}{s^2+2s+2}$

By anti-transforming the fraction we get
$U(s)=ccL[f(x)](s)*ccL(e^(-x)sinx)(s)$

Eventually,
$ccL[u(x)](s)=ccL[f(x)^**e^(-x)sinx](s)$
$u(x)=f(x)^**e^(-x)sinx$

Is it correct? I do hope so, but I'm not so sure. Please let me know, as soon as you can. Thanks a lot.

Thanks.
Paolo