# Math Help - Laplace Transform Involving Initial Value Problem

1. So, is the answer i posted correct?

2. 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

3. 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

4. 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 $Y(0) = -5.$

5. 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

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

7. Originally Posted by Belowzero78
So, the function f(t) i posted is correct given the intervals of t?