# Laplace transforms

• Nov 7th 2006, 12:30 PM
macabre
Laplace transforms
Ok, so I've got the question:

Use laplace transforms to find x(t) when x and y satisfy the differential equations:

dx/dt -dy/dt -2x = 0

and

dx/dt +dy/dt -2y = t

With x = 1, y = 2 at t = 0

So far i've taken laplace transforms of each side and substituted one into the other to get a single equation for Laplace(x)

and got to the point where

Laplace(x) = {3*(1/s)*[(s-1)^2/(s-1)^2 + 1]} + [3* 1/(s-1)^2 + 1] -{15/6 *(1/s)*(1/(s-1)^2 + 1)}

So now i presume that i just inverse the laplace transforms that i have gotten, but how do you invert the parts where there are two transforms multiplied together such as the 1st and 3rd terms in the above?

is it just a case of : Laplace^-1 {3*(1/s)*[(s-1)^2/(s-1)^2 + 1]} is

simply the product of the individual transforms once they have been inverted ie

3 * 1 * exp(t) * cos(t)

OR is there some other way that i am not aware of?
• Nov 7th 2006, 07:49 PM
ThePerfectHacker
Quote:

Originally Posted by macabre
Ok, so I've got the question:

Use laplace transforms to find x(t) when x and y satisfy the differential equations:

dx/dt -dy/dt -2x = 0

and

dx/dt +dy/dt -2y = t

With x = 1, y = 2 at t = 0

Let, the functional transform be,
$\displaystyle \int_0^{\infty}e^{-st}x(t)dt=F(s)$
$\displaystyle \int_0^{\infty}e^{-st}y(t)dt=G(s)$
Then the Laplace Transforms of the derivatives are respectively:
$\displaystyle -x(0)+sF(s)=-1+sF(s)$
$\displaystyle -y(0)+sG(s)=-2+sG(s)$
Take Laplace Transform of both sides.
Thus, we have, from the system of differencial equations,
$\displaystyle -1+sF(s)+2-sG(s)-2F(s)=0$
$\displaystyle -1+sF(s)-2+sG(s)-2G(s)=\frac{1}{s}$
Thus,
$\displaystyle F(s)(s-2)=sG(s)-1$
Thus, (note $\displaystyle s>2$)
$\displaystyle F(s)=\frac{sG(s)-1}{s-2}$
Substitute into second functional equation,
$\displaystyle -3+\frac{s^2G(s)-s}{s-2}+sG(s)-2G(s)=\frac{1}{s}$
Multiply, to clear denominator,
$\displaystyle -3(s-2)+s^2G(s)-s+s(s-2)G(s)-2(s-2)G(s)=\frac{s-2}{s}$
Thus,
$\displaystyle G(s)[s^2+s(s-2)-2(s-2)]-3(s-2)-s=\frac{s-2}{s}$
Thus,
$\displaystyle G(s)[s^2+s^2-2s-2s+4]-3s+4-s=\frac{s-2}{s}$
Thus,
$\displaystyle G(s)[2s^2-4s+4]-4s+4=\frac{s-2}{s}$
Thus,
$\displaystyle G(s)[2s^2-4s+4]=\frac{s-2}{s}+4s-4$
Thus,
$\displaystyle G(s)[2s^2-4s+4]=\frac{s-2+4s^2-4s}{s}$
Thus,
$\displaystyle G(s)=\frac{4s^2-3s-2}{s(2s^2-4s+4)}$
Now I think there is something wrong with your problem because the procedure now is to decompose after factorization but the factorization is irrational, unlikely that they will give that as a homework problem.
• Nov 8th 2006, 08:03 AM
TD!
Quote:

Originally Posted by ThePerfectHacker
Thus, we have, from the system of differencial equations,
$\displaystyle -1+sF(s)+2-sG(s)-2F(s)=0$
$\displaystyle -1+sF(s)-2+sG(s)-2G(s)=\frac{1}{s}$

The Laplace transform of t is 1/sē, not 1/s. We have:

$\displaystyle \mathcal{L}\left\{ {x' - y' - 2x} \right\} = sX - x_0 - sY + y_0 - 2X = \left( {s - 2} \right)X - sY - 3$

and

$\displaystyle \mathcal{L}\left\{ {x' - y' - 2y - t} \right\} = sX - x_0 - sY + y_0 - 2Y - s^{ - 2} = sX - \left( {s + 2} \right)Y - s^2 - 3$

So this yield the following system of equations in X and Y:

$\displaystyle \left\{ \begin{array}{l} \left( {s - 2} \right)X - sY = 3 \\ sX - \left( {s + 2} \right)Y = s^2 + 3 \\ \end{array} \right.$

With solutions:

$\displaystyle X = \frac{1}{4}s - \frac{3}{2} \; \wedge \; Y = \frac{1}{4}s - \frac{1}{2}s^2 - \frac{3}{2}$

Now $\displaystyle x = \mathcal{L}^{-1}(X) \; \wedge \; y = \mathcal{L}^{-1}(Y)$

I doubt you'll have to work with Dirac-impulses though, so I'm assuming you won't have to take those constant terms into account.
If so, then you get the easy solution x(t) = 1/4 and y(t) = 1/4 - t/2 which indeed satisties the system of DE's; if not: you'll get $\displaystyle \delta{(t)}$ as well.
• Nov 8th 2006, 02:20 PM
macabre
TD! - the 2nd equation is x' + y' - 2y = t not x'-y' -2x = t

Also, how have you gone from having s^-2 to s^2 in your working?
• Nov 8th 2006, 11:09 PM
TD!
About the sē, I just forgot to put the minus up there, the solution I gave holds for 1/sē, not sē.
In the second equation I did have -2y instead of -2x, but I got the sign of y' wrong, apparently.

Let me fix my mistakes :)

$\displaystyle \mathcal{L}\left\{ {x' - y' - 2x} \right\} = sX - x_0 - sY + y_0 - 2X = \left( {s - 2} \right)X - sY + 1$

$\displaystyle \mathcal{L}\left\{ {x' + y' - 2y - t} \right\} = sX - x_0 + sY - y_0 - 2Y - s^{ - 2} = sX + \left( {s - 2} \right)Y - s^2 - 3$

This gives the system:

$\displaystyle \left\{ \begin{array}{l} \left( {s - 2} \right)X - sY = -1 \\ sX + \left( {s - 2} \right)Y = s^2 + 3 \\ \end{array} \right.$

Unfortunately, the solution is a bit more complicated for this one:

$\displaystyle X = \frac{{2s^2 + 2s + 1}}{{2s\left( {s^2 - 2s + 2} \right)}} \; \wedge \; Y = \frac{{4s^3 - 6s^2 + s - 2}}{{2s^2 \left( {s^2 - 2s + 2} \right)}}$

I assume you know how to do a partial fraction decomposition, this gives:

$\displaystyle \begin{array}{*{20}l} X & { = \frac{{3s}}{{4\left( {s^2 - 2s + 2} \right)}} + \frac{3}{{2\left( {s^2 - 2s + 2} \right)}} + \frac{1}{{4s}}} \\ \\ {} & { = \frac{3}{4}\frac{{s - 1}}{{\left( {s - 1} \right)^2 + 1}} + \frac{9}{4}\frac{1}{{\left( {s - 1} \right)^2 + 1}} + \frac{1}{{4s}}} \end{array}$

$\displaystyle \begin{array}{*{20}l} Y & { = \frac{{9s}}{{4\left( {s^2 - 2s + 2} \right)}} - \frac{3}{{\left( {s^2 - 2s + 2} \right)}} - \frac{1}{{2s^2 }} - \frac{1}{{4s}}} \\ \\ {} & { = \frac{9}{4}\frac{{s - 1}}{{\left( {s - 1} \right)^2 + 1}} - \frac{3}{4}\frac{1}{{\left( {s - 1} \right)^2 + 1}} - \frac{1}{{2s^2 }} - \frac{1}{{4s}}} \end{array}$

In the last line for X and Y, I rewrote so they're exactly in the standard form as you can look them up in a table.
To get x(t) and y(t) out of X(s) and Y(s), you now do the inverse Laplace transform as follows:

$\displaystyle \begin{array}{*{20}l} x & {= \mathcal{L}^{ - 1} \left\{ X \right\}} \\ \\ {} & { = \frac{3}{4}\mathcal{L}^{ - 1} \left\{ {\frac{{s - 1}}{{\left( {s - 1} \right)^2 + 1}}} \right\} + \frac{9}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{{\left( {s - 1} \right)^2 + 1}}} \right\} + \frac{1}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{s}} \right\}} \\ \\ {} & { = \frac{3}{4}e^t \cos t + \frac{9}{4}e^t \sin t + \frac{1}{4}} \end{array}$

$\displaystyle \begin{array}{*{20}l} y & {= \mathcal{L}^{ - 1} \left\{ Y \right\}} \\ \\ {} & { = \frac{9}{4}\mathcal{L}^{ - 1} \left\{ {\frac{{s - 1}}{{\left( {s - 1} \right)^2 + 1}}} \right\} - \frac{3}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{{\left( {s - 1} \right)^2 + 1}}} \right\} - \frac{1}{2}\mathcal{L}^{ - 1} \left\{ {\frac{1}{{s^2 }}} \right\} - \frac{1}{4}\mathcal{L}^{ - 1} \left\{ {\frac{1}{s}} \right\}} \\ \\ {} & { = \frac{9}{4}e^t \cos t - \frac{3}{4}e^t \sin t - \frac{t}{2} - \frac{1}{4}} \end{array}$

I hope I didn't make any mistakes this time, but I checked the solution with the 'classic method' and it should be ok:

$\displaystyle \left\{ \begin{array}{l} x\left( t \right) = \frac{1}{4}\left( {3e^t \cos t + 9e^t \sin t + 1} \right) \\ \\ y\left( t \right) = \frac{1}{4}\left( {9e^t \cos t - 3e^t \sin t - 2t - 1} \right) \end{array} \right.$