# Thread: Laplace Transform Involving Initial Value Problem

1. ## 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!

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

Regards

Fernando Revilla

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

4. 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

5. Would this be the correct answer?

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

6. 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

7. 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

8. 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.

9. 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

10. 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!

11. 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

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

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

13. Originally Posted by CaptainBlack
$\mathcal{L}(f'(t))=F(s)-f(0)$
you missed an "s" there.

14. Originally Posted by Krizalid
you missed an "s" there.
Fixed

CB

15. 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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