Laplace Transform Involving Initial Value Problem

**Belowzero78** · November 23rd 2010, 06:31 PM

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!

**FernandoRevilla** · November 23rd 2010, 09:14 PM

If $y(t)=-5e^{-t},\;t\in [0,7)$ then, $y'(t)+y(t)=0\neq f(t)$ .

Regards

Fernando Revilla

**Belowzero78** · November 23rd 2010, 09:17 PM

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

**FernandoRevilla** · November 23rd 2010, 09:44 PM

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

**Belowzero78** · November 24th 2010, 07:46 PM

Would this be the correct answer?

$f(t)=\left\{\begin{array}{cc}-8e^{-t}&\mbox{ for } 0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.$

**FernandoRevilla** · November 24th 2010, 09:42 PM

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

**CaptainBlack** · November 24th 2010, 11:39 PM

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

**Belowzero78** · November 25th 2010, 05:53 AM

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.

**CaptainBlack** · November 25th 2010, 09:47 PM

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

**Belowzero78** · November 25th 2010, 09:57 PM

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!

**CaptainBlack** · November 26th 2010, 12:13 AM

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

**Belowzero78** · November 26th 2010, 05:16 PM

This is my final answer now. Please confirm it is correct. Thanks!

$f(t)=\left\{\begin{array}{cc}3 - 8e^{-t}&\mbox{ for } 0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.$

**Krizalid** · November 26th 2010, 06:51 PM

Originally Posted by CaptainBlack

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

you missed an "s" there.

**CaptainBlack** · November 26th 2010, 07:51 PM

Originally Posted by Krizalid

you missed an "s" there.

Fixed

CB

**Prove It** · November 26th 2010, 07:58 PM

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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