Laplace Transform Involving Initial Value Problem

**Belowzero78** · November 26th 2010, 08:23 PM

So, is the answer i posted correct?

**CaptainBlack** · November 27th 2010, 12:46 AM

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

**CaptainBlack** · November 27th 2010, 12:53 AM

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

**Belowzero78** · November 27th 2010, 07:57 AM

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

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

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 $Y(0) = -5.$

**CaptainBlack** · November 27th 2010, 08:15 AM

Originally Posted by Belowzero78

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

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

**Belowzero78** · November 27th 2010, 08:16 AM

So, the function f(t) i posted is correct given the intervals of t?

**CaptainBlack** · November 28th 2010, 10:26 AM

Originally Posted by Belowzero78

So, the function f(t) i posted is correct given the intervals of t?

It has already been confirmed.

CB

Thread: Laplace Transform Involving Initial Value Problem

LinkBack

Thread Tools

Display

Similar Math Help Forum Discussions

Including initial conditions in Laplace Transform MATLAB

Laplace transform involving Heaviside functions

Laplace Transforms to solve Initial-Value problem

Laplace transform to solve the initial value

Initial Value Problem 3rd order DE - using laplace