# Laplace Transform Involving Initial Value Problem

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• 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?

It has already been confirmed.

CB
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