# Laplace Transform Involving Initial Value Problem

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Nov 26th 2010, 08:23 PM
Belowzero78
So, is the answer i posted correct?
• Nov 27th 2010, 12:46 AM
CaptainBlack
Quote:

Originally Posted by Prove It
Are Laplace Transforms really necessary here?

Not to solve the ODE. But to do the question yes, it asks the student to use Laplace transforms.

CB
• Nov 27th 2010, 12:53 AM
CaptainBlack
Quote:

Originally Posted by Belowzero78
So, is the answer i posted correct?

You can check this yourself. There are two requirements that the solution must full fill, it must satisfy the differential equation tofether with the initial condition and be continuous at the knot (the point where the forcing function changes its' form).

You solution appears to meet these requirements.

CB
• Nov 27th 2010, 07:57 AM
Belowzero78
Quote:

Originally Posted by CaptainBlack
You can check this yourself. There are two requirements that the solution must full fill, it must satisfy the differential equation tofether with the initial condition and be continuous at the knot (the point where the forcing function changes its' form).

You solution appears to meet these requirements except it does not satisfy the initial condition.

CB

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

Why doesn't this satisfy the initial condition? I'll show my work below.

$\displaystyle{Y(s) = \frac{-5}{s+1} -\frac{6e^{-7s}}{s(s+1)} + \frac{3}{s(s+1)}}$

$\displaystyle{Y(s) = \frac{-5}{s+1} +(\frac{-6}{s} + \frac{6}{s+1})e^{-7s} + \frac{3}{s} - \frac{3}{s+1}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = \mathfrak{L^{-1}}\{\frac{-5}{s+1}\}(t) +\mathfrak{L^{-1}}\{(\frac{-6}{s} + \frac{6}{s+1})e^{-7s}\}(t) + \mathfrak{L^{-1}}\{\frac{3}{s}\}(t) - \mathfrak{L^{-1}}\{\frac{3}{s+1})\}(t)}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = -5e^{-t} + \mathfrak{L^{-1}}\{\frac{-6}{s} + \frac{6}{s+1}\}(t - 7)\cdot U_7(t) + 3 - 3e^{-t}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{-t})|_{t = t - 7} \cdot U_7(t)$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{7 - t}) \cdot U_7(t)$

So this is what gives me the answer. I still am not sure why it doesn't satisfy the initial conditions given the fact I used it right at the start when $Y(0) = -5.$
• Nov 27th 2010, 08:15 AM
CaptainBlack
Quote:

Originally Posted by Belowzero78
$f(t)=\left\{\begin{array}{cc}3 - 8e^{-t}&\mbox{ for }
0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.$

Why doesn't this satisfy the initial condition? I'll show my work below.

$\displaystyle{Y(s) = \frac{-5}{s+1} -\frac{6e^{-7s}}{s(s+1)} + \frac{3}{s(s+1)}}$

$\displaystyle{Y(s) = \frac{-5}{s+1} +(\frac{-6}{s} + \frac{6}{s+1})e^{-7s} + \frac{3}{s} - \frac{3}{s+1}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = \mathfrak{L^{-1}}\{\frac{-5}{s+1}\}(t) +\mathfrak{L^{-1}}\{(\frac{-6}{s} + \frac{6}{s+1})e^{-7s}\}(t) + \mathfrak{L^{-1}}\{\frac{3}{s}\}(t) - \mathfrak{L^{-1}}\{\frac{3}{s+1})\}(t)}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = -5e^{-t} + \mathfrak{L^{-1}}\{\frac{-6}{s} + \frac{6}{s+1}\}(t - 7)\cdot U_7(t) + 3 - 3e^{-t}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{-t})|_{t = t - 7} \cdot U_7(t)$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{7 - t}) \cdot U_7(t)$

So this is what gives me the answer. I still am not sure why it doesn't satisfy the initial conditions given the fact I used it right at the start when $Y(0) = -5.$

Arithmetic error on my part.

To check your solution does not require that we look at your work, though it is nice to see it finally.

CB
• Nov 27th 2010, 08:16 AM
Belowzero78
So, the function f(t) i posted is correct given the intervals of t?
• Nov 28th 2010, 10:26 AM
CaptainBlack
Quote:

Originally Posted by Belowzero78
So, the function f(t) i posted is correct given the intervals of t?
The content structure of the provided text is a discussion thread from a math forum, where a user is asking for verification of their solution to a problem involving Laplace transforms and initial value problems.
no_question

the text does not contain a mathematical question that is being posed for an answer; rather, it is a discussion where a user is seeking confirmation of their solution. the mathematical problem and its solution are already provided, and the discussion revolves around verifying the correctness of the solution.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Laplace Transform Involving Initial Value Problem

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• Nov 26th 2010, 08:23 PM
Belowzero78
So, is the answer i posted correct?
• Nov 27th 2010, 12:46 AM
CaptainBlack
Quote:

Originally Posted by Prove It
Are Laplace Transforms really necessary here?

Not to solve the ODE. But to do the question yes, it asks the student to use Laplace transforms.

CB
• Nov 27th 2010, 12:53 AM
CaptainBlack
Quote:

Originally Posted by Belowzero78
So, is the answer i posted correct?

You can check this yourself. There are two requirements that the solution must full fill, it must satisfy the differential equation tofether with the initial condition and be continuous at the knot (the point where the forcing function changes its' form).

You solution appears to meet these requirements.

CB
• Nov 27th 2010, 07:57 AM
Belowzero78
Quote:

Originally Posted by CaptainBlack
You can check this yourself. There are two requirements that the solution must full fill, it must satisfy the differential equation tofether with the initial condition and be continuous at the knot (the point where the forcing function changes its' form).

You solution appears to meet these requirements except it does not satisfy the initial condition.

CB

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

Why doesn't this satisfy the initial condition? I'll show my work below.

$\displaystyle{Y(s) = \frac{-5}{s+1} -\frac{6e^{-7s}}{s(s+1)} + \frac{3}{s(s+1)}}$

$\displaystyle{Y(s) = \frac{-5}{s+1} +(\frac{-6}{s} + \frac{6}{s+1})e^{-7s} + \frac{3}{s} - \frac{3}{s+1}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = \mathfrak{L^{-1}}\{\frac{-5}{s+1}\}(t) +\mathfrak{L^{-1}}\{(\frac{-6}{s} + \frac{6}{s+1})e^{-7s}\}(t) + \mathfrak{L^{-1}}\{\frac{3}{s}\}(t) - \mathfrak{L^{-1}}\{\frac{3}{s+1})\}(t)}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = -5e^{-t} + \mathfrak{L^{-1}}\{\frac{-6}{s} + \frac{6}{s+1}\}(t - 7)\cdot U_7(t) + 3 - 3e^{-t}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{-t})|_{t = t - 7} \cdot U_7(t)$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{7 - t}) \cdot U_7(t)$

So this is what gives me the answer. I still am not sure why it doesn't satisfy the initial conditions given the fact I used it right at the start when $Y(0) = -5.$
• Nov 27th 2010, 08:15 AM
CaptainBlack
Quote:

Originally Posted by Belowzero78
$f(t)=\left\{\begin{array}{cc}3 - 8e^{-t}&\mbox{ for }
0\leq t <7\\-8e^{-t} + 6e^{7-t} - 3 & \mbox{ for } t\geq7\end{array}\right.$

Why doesn't this satisfy the initial condition? I'll show my work below.

$\displaystyle{Y(s) = \frac{-5}{s+1} -\frac{6e^{-7s}}{s(s+1)} + \frac{3}{s(s+1)}}$

$\displaystyle{Y(s) = \frac{-5}{s+1} +(\frac{-6}{s} + \frac{6}{s+1})e^{-7s} + \frac{3}{s} - \frac{3}{s+1}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = \mathfrak{L^{-1}}\{\frac{-5}{s+1}\}(t) +\mathfrak{L^{-1}}\{(\frac{-6}{s} + \frac{6}{s+1})e^{-7s}\}(t) + \mathfrak{L^{-1}}\{\frac{3}{s}\}(t) - \mathfrak{L^{-1}}\{\frac{3}{s+1})\}(t)}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = -5e^{-t} + \mathfrak{L^{-1}}\{\frac{-6}{s} + \frac{6}{s+1}\}(t - 7)\cdot U_7(t) + 3 - 3e^{-t}}$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{-t})|_{t = t - 7} \cdot U_7(t)$

$\displaystyle{\mathfrak{L^{-1}}\{Y(s)\}(t) = 3 - 8e^{-t} + (-6 + 6e^{7 - t}) \cdot U_7(t)$

So this is what gives me the answer. I still am not sure why it doesn't satisfy the initial conditions given the fact I used it right at the start when $Y(0) = -5.$

Arithmetic error on my part.

To check your solution does not require that we look at your work, though it is nice to see it finally.

CB
• Nov 27th 2010, 08:16 AM
Belowzero78
So, the function f(t) i posted is correct given the intervals of t?
• Nov 28th 2010, 10:26 AM
CaptainBlack
Quote:

Originally Posted by Belowzero78
So, the function f(t) i posted is correct given the intervals of t?
user
The content structure of the provided text is a discussion thread from a math forum, where a user is asking for verification of their solution to a problem involving Laplace transforms and initial value problems.

TYPE: NO_QUESTION

The text does not contain a mathematical question that is being posed for an answer; rather, it is a discussion where a user is seeking confirmation of their solution. The mathematical problem and its solution are already provided, and the discussion revolves around verifying the correctness of the solution.