# show that y'' − y = F(t) can be written...

• Apr 27th 2011, 03:37 AM
poirot
show that y'' − y = F(t) can be written...
Show that y'' − y = F(t) can be written as y= Ae^t+Be^-t + \int sinh(t-\tau )F\tau d\tau

Obviously, the first is the complementary solution which is easily obtained. The second part I have no idea about. Tried using a variation of parameters but how to get that integral I don't know.
• Apr 27th 2011, 04:46 AM
Ackbeet
Why don't you just differentiate the y there twice, plug into the DE, and see if you get equality?
• Apr 27th 2011, 05:43 AM
poirot
Ok I've tried that but I'm having difficulty integrating with respect to tau as I use parts but that just creates another difficult integral and then i will have to differentiate all that twice
• Apr 27th 2011, 05:59 AM
Ackbeet
My suggestion was to differentiate, not integrate. That is, you have

$y(t)=Ae^{t}+Be^{-t}+\int_{0}^{t}\sinh(t-\tau)F(\tau)\,d\tau\implies$

$\dot{y}(t)=Ae^{t}-Be^{-t}+\overbrace{\int_{0}^{t}\cosh(t-\tau)F(\tau)\,d\tau+\underbrace{\sinh(t-t)F(t)\cdot 1}_{=0}-\underbrace{\sinh(t-0)F(0)\cdot 0}_{=0}}^{\text{Use Leibniz's Integral Rule here}}$

$=Ae^{t}-Be^{-t}+\int_{0}^{t}\cosh(t-\tau)F(\tau)\,d\tau.$

Here's a link to Leibniz's Integral Rule. You'll need to differentiate again.
• Apr 27th 2011, 06:40 AM
poirot
when I differentiate again I just get y so my final answer would be 0. partial d/dt(cosh(t-tau)) = sinh(t-tau)
• Apr 27th 2011, 06:43 AM
Ackbeet
No, you should not get just $y$. The second time around, the two boundary terms from the Leibniz integral rule are not both zero. Check the rule a little more carefully.
• Apr 27th 2011, 07:00 AM
topsquark
Quote:

Originally Posted by Ackbeet
No, you should not get just $y$. The second time around, the two boundary terms from the Leibniz integral rule are not both zero. Check the rule a little more carefully.

A quick check-in, then I'll get out of the way of the main discussion. The Leibniz rule. That is essentially the Fundamental Theorem mixed in with the chain rule, right? (I know that's only a brief description, I can fill in the details on my own if that is where it comes from.)

Thanks!
-Dan
• Apr 27th 2011, 07:01 AM
Ackbeet
Quote:

Originally Posted by topsquark
A quick check-in, then I'll get out of the way of the main discussion. The Leibniz rule. That is essentially the Fundamental Theorem mixed in with the chain rule, right? (I know that's only a brief description, I can fill in the details on my own if that is where it comes from.)

Thanks!
-Dan

You are essentially correct, yes.
• Apr 27th 2011, 07:01 AM
poirot
I have got the answer. Thanks a lot for your help. Just one question: when you have multiplied the boundary terms by 1 and 0 respectively, does this represent d/dt(t) and d/dt(0) respectively.
• Apr 27th 2011, 07:02 AM
Ackbeet
Quote:

Originally Posted by poirot
I have got the answers. Thanks a lot for your help. Just one question: when you have multiplied the boundary terms by 1 and 0 respectively, does this represent d/dt(t) and d/dt(0) respectively.

Exactly right. You're welcome!
• Apr 27th 2011, 10:19 PM
chisigma
Quote:

Originally Posted by poirot
Show that y'' − y = F(t) can be written as y= Ae^t+Be^-t + \int sinh(t-\tau )F\tau d\tau

Obviously, the first is the complementary solution which is easily obtained. The second part I have no idea about. Tried using a variation of parameters but how to get that integral I don't know.

In terms of Laplace Transform, supposing y(0)=y'(0)=0, the DE is written as...

http://quicklatex.com/cache3/ql_4d7f...b6883df_l3.png (1)

... so that y(t) is derived as follows...

http://quicklatex.com/cache3/ql_dcb8...c71bac7_l3.png

Kind regards

$\chi$ $\sigma$