# Laplace Transform Involving Initial Value Problem

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• Nov 23rd 2010, 06:31 PM
Belowzero78
Laplace Transform Involving Initial Value Problem
Question: Use Laplace transforms to solve the initial value problem $\frac{dy}{dt} + y = f(t)$, where $f(t)=\left\{\begin{array}{cc}3&\mbox{ for }
0\leq t <7\\-3 & \mbox{ for } t\geq7\end{array}\right.$
, and $y(0) = -5$.

My Solution is: $f(t)=\left\{\begin{array}{cc}-5e^{-t}&\mbox{ for }
0\leq t <7\\-5e^{-t} + 6e^{7-t} - 6 & \mbox{ for } t\geq7\end{array}\right.$

I am confident for the first interval $0\leq t < 7$ that the answer is correct, but since i got the incorrect answer for the 2nd part, i was wondering since at $t\geq7$, when $t = 7$, the function $y(t)$ is still continuous, but not differentiable. What should be the correct answer and whoever contributes, could you please explain why?

Hope to hear from you all soon!
• Nov 23rd 2010, 09:14 PM
FernandoRevilla
If $y(t)=-5e^{-t},\;t\in [0,7)$ then, $y'(t)+y(t)=0\neq f(t)$ .

Regards

Fernando Revilla
• Nov 23rd 2010, 09:17 PM
Belowzero78
So what are you trying to say exactly? I'm not catching it. Which part is wrong?
• Nov 23rd 2010, 09:44 PM
FernandoRevilla
Quote:

Originally Posted by Belowzero78
So what are you trying to say exactly? I'm not catching it. Which part is wrong?

The $y(t)$ function does not satisfy the differential equation, so there must be some mistake.

Regards.

Fernando Revilla
• Nov 24th 2010, 07:46 PM
Belowzero78
Would this be the correct answer?

$f(t)=\left\{\begin{array}{cc}-8e^{-t}&\mbox{ for }
0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.$
• Nov 24th 2010, 09:42 PM
FernandoRevilla
Take into account that your solution does not satisfy the initial condition $y(0)=-5$ . You should obtain $y(t)=3-8e^{-t}\;(x\in[0,7))$ .

Regards.

Fernando Revilla
• Nov 24th 2010, 11:39 PM
CaptainBlack
Quote:

Originally Posted by Belowzero78
Question: Use Laplace transforms to solve the initial value problem $\frac{dy}{dt} + y = f(t)$, where $f(t)=\left\{\begin{array}{cc}3&\mbox{ for }
0\leq t <7\\-3 & \mbox{ for } t\geq7\end{array}\right.$
, and $y(0) = -5$.

My Solution is: $f(t)=\left\{\begin{array}{cc}-5e^{-t}&\mbox{ for }
0\leq t <7\\-5e^{-t} + 6e^{7-t} - 6 & \mbox{ for } t\geq7\end{array}\right.$

I am confident for the first interval $0\leq t < 7$ that the answer is correct, but since i got the incorrect answer for the 2nd part, i was wondering since at $t\geq7$, when $t = 7$, the function $y(t)$ is still continuous, but not differentiable. What should be the correct answer and whoever contributes, could you please explain why?

Hope to hear from you all soon!

rewrite $f(t)$ as

$f(t)=-5e^{-t} u(t) + (6e^{7-t}-6)u(t-7)$

Now take the Laplace transform of the ODE and what do you get?

CB
• Nov 25th 2010, 05:53 AM
Belowzero78
Ok, thanks for the help so far. Now, as far as for your reply, how do you which answer to put in for the lnterval or boundary condition? I think I got the mathematics down, but of how to answer the question given the initial conditions.
• Nov 25th 2010, 09:47 PM
CaptainBlack
Quote:

Originally Posted by Belowzero78
Ok, thanks for the help so far. Now, as far as for your reply, how do you which answer to put in for the lnterval or boundary condition? I think I got the mathematics down, but of how to answer the question given the initial conditions.

When you take the LT of the left hand side the initial conditions will be used since:

$\mathcal{L}(f'(t))=sF(s)-f(0)$

where $F(s)=\mathcal{L} (f(t))$

CB
• Nov 25th 2010, 09:57 PM
Belowzero78
So what is the final answer? I really want to know because I've been stuck for 2 days now and make sure I get it correct since this for an online Assignment. Thanks alot to everyone!
• Nov 26th 2010, 12:13 AM
CaptainBlack
Quote:

Originally Posted by Belowzero78
So what is the final answer? I really want to know because I've been stuck for 2 days now and make sure I get it correct since this for an online Assignment. Thanks alot to everyone!

No, what you do is post the result of Laplace transforming the equation and we will see if that is ok then see how you are inverting the result.

CB
• Nov 26th 2010, 05:16 PM
Belowzero78
This is my final answer now. Please confirm it is correct. Thanks!

$f(t)=\left\{\begin{array}{cc}3 - 8e^{-t}&\mbox{ for }
0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.$
• Nov 26th 2010, 06:51 PM
Krizalid
Quote:

Originally Posted by CaptainBlack
$\mathcal{L}(f'(t))=F(s)-f(0)$

you missed an "s" there. :D
• Nov 26th 2010, 07:51 PM
CaptainBlack
Quote:

Originally Posted by Krizalid
you missed an "s" there. :D

Fixed

CB
• Nov 26th 2010, 07:58 PM
Prove It
Are Laplace Transforms really necessary here?

Case 1: $\displaystyle \frac{dy}{dt} + y = 3$

The Integrating Factor is $\displaystyle e^{\int{1\,dt}} = e^t$, so multiplying through gives

$\displaystyle e^t\,\frac{dy}{dt} + e^ty = 3e^t$

$\displaystyle \frac{d}{dt}(e^ty)= 3e^t$

$\displaystyle e^ty = \int{3e^t\,dt}$

$\displaystyle e^ty = 3e^t + C$

$\displaystyle y = 3 + Ce^{-t}$.

Using the initial condition $\displaystyle y(0)=-5$, we find $\displaystyle -5 = 3 + Ce^{-0}$, which means $\displaystyle C = -8$.

Thus the first solution is $\displaystyle y= 3 - 8e^{-t}$.

Use a similar process for Case 2, when $\displaystyle f(t) = -3$.
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