System of DE's: Laplace transfomation

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April 28th 2010, 01:19 AM
Kbotz

System of DE's: Laplace transfomation

Using the Laplace transform, find the functions x(t) and y(t) that satisfy

x=y=0 at t=0 and

dx/dt + 2(dy/dt) + x = t
-dx/dt - dy/dt + y =1
April 28th 2010, 05:02 AM
mr fantastic

Quote:

Originally Posted by Kbotz

Using the Laplace transform, find the functions x(t) and y(t) that satisfy

x=y=0 at t=0 and

dx/dt + 2(dy/dt) + x = t
-dx/dt - dy/dt + y =1

What have you tried? Where do you get stuck?
April 28th 2010, 10:23 AM
Kbotz

i transformed each one of the variables.
so for the 1st equation i got

sX+2sY = 1/s^2 ......................... eq 3

and transformation of the second equation
-sX-sY = 1/s.............................. eq 4

then eliminated X from both equations by multiplying (eq 3) by -1, and taking eq 4 from it.
Then made Y the subject

Y = 1+s/s^3

I want to know if the previous steps are right, and the approach to the problem.

Thanks Fantastic.
April 28th 2010, 06:39 PM
mr fantastic

Quote:

Originally Posted by Kbotz

i transformed each one of the variables.
so for the 1st equation i got

sX+2sY = 1/s^2 ......................... eq 3

and transformation of the second equation
-sX-sY = 1/s.............................. eq 4

[snip]

What about the Laplace transform of x and y in the original DE's?

The correct equations are:

sX + 2sY + X = 1/s^2 ......................... eq 3

-sX - sY + Y = 1/s.............................. eq 4
April 28th 2010, 08:13 PM
Kbotz

oh right, thanks
i will work on it from here and post back if i need more help!
thanks again for the heads up mr.fantastic!
May 2nd 2010, 09:35 PM
Kbotz

hey,
i ran into more trouble doing this problem.
i have no idea whether my answer is right.

transformed equations as in earlier post
$ sX + 2sY + X = 1/s^2$ eqn (1)

$-sX - sY + Y= 1/s$ eqn (2)

Multiply equation 1 by $s^2$ to remove the $s^2$ on the RHS.

$ $ $s^3X + 2s^3Y + s^2X = 1$ eqn (1')

Multiply equation 2 by s to remove the 's' on the RHS.
$ $ $-s^2x - s^2Y + sY = 1$ eqn (2')

equation 1' & 2' commonly factorized in terms of X
$ X(s^3+s^2) + 2s^3Y = 1$ .... eqn(1')
$X(-s^2) - s^2Y + sY = 1$ .....eqn(2')

To eliminate X from the pair of equations; eqn 2' is multiplied by (-s-1)

After further simplification

$X(s^3 + s^2) + s^3Y - sY = -s-1$ .....eqn(2'')

Now X can be eliminated, so subtract eqn (2'') from eqn (1')

I don't know if this is right, after subtraction i got:

$s^3Y + sY = 2-s Y(s^3+s) = 2-s$

Make Y the subject

$Y = 2-s/(s^3+s)$ , which is the same as $Y = 2-s/(s*(s^2+1))$

Now using partial fractions

$2-s = A/s + B/(s^2+1)$

cross multiply

$2-s = A(s^2+1) + B(s)$

when s = 0, B=0 and A = 2

then when s =1, substitute in A =2 to find out B

$s = 1, B = -3$

Taking the inverse laplace transform of the partial fraction:

$Y = 2/s - 3/(s^2+1)$

$y(t) = 2 - 3sin(t)$

Please check my steps, i hope it makes sense.
And i have no idea how to get x(t), i keep coming to really complex equations.
May 3rd 2010, 12:48 AM
mr fantastic

Quote:

Originally Posted by Kbotz

hey,
i ran into more trouble doing this problem.
i have no idea whether my answer is right.

transformed equations as in earlier post
$ sX + 2sY + X = 1/s^2$ eqn (1)

$-sX - sY + Y= 1/s$ eqn (2)

Multiply equation 1 by $s^2$ to remove the $s^2$ on the RHS.

$ s^3X + 2s^3Y + s^2X = 1$ eqn (1')

Multiply equation 2 by s to remove the 's' on the RHS.
$ -s^2x - s^2Y + sY = 1$ eqn (2')

equation 1' & 2' commonly factorized in terms of X
$ X(s^3+s^2) + 2s^3Y = 1$ .... eqn(1')
$X(-s^2) - s^2Y + sY = 1$ .....eqn(2')

To eliminate X from the pair of equations; eqn 2' is multiplied by (-s-1)

After further simplification

$X(s^3 + s^2) + s^3Y - sY = -s-1$ .....eqn(2'')

Now X can be eliminated, so subtract eqn (2'') from eqn (1')

I don't know if this is right, after subtraction i got:

$s^3Y + sY = 2-s Y(s^3+s) = 2-s$

Make Y the subject

$Y = 2-s/(s^3+s)$ , which is the same as $Y = 2-s/(s*(s^2+1))$

Now using partial fractions

$2-s = A/s + B/(s^2+1)$

cross multiply

$2-s = A(s^2+1) + B(s)$

when s = 0, B=0 and A = 2

then when s =1, substitute in A =2 to find out B

$s = 1, B = -3$

Taking the inverse laplace transform of the partial fraction:

$Y = 2/s - 3/(s^2+1)$

$y(t) = 2 - 3sin(t)$

Please check my steps, i hope it makes sense.
And i have no idea how to get x(t), i keep coming to really complex equations.

I get:

$x = - \frac{(2s^2 + s - 1)}{s^2 (s^2 + 1)} = \frac{s}{s^2 + 1} - \frac{3}{s^2 + 1} - \frac{1}{s} + \frac{1}{s^2}$

$y = \frac{s+2}{s (s^2 + 1)} = \frac{-2s}{s^2 + 1} + \frac{1}{s^2 + 1} + \frac{2}{s}$ .

You just have to use basic algebra and solve the equations simultaneously for x and y and then express each solution in partial fraction form.
May 3rd 2010, 12:54 AM
Kbotz

Quote:

Originally Posted by mr fantastic

I get:

$x = - \frac{(2s^2 + s - 1)}{s^2 (s^2 + 1)} = \frac{s}{s^2 + 1} - \frac{3}{s^2 + 1} - \frac{1}{s} + \frac{1}{s^2}$

$y = \frac{s+2}{s (s^2 + 1)} = \frac{-2s}{s^2 + 1} + \frac{1}{s^2 + 1} + \frac{2}{s}$ .

You just have to use basic algebra and solve the equations simultaneously for x and y and then express each solution in partial fraction form.

i'm a little unsure of how you got s+2 for the y equation, but thanks for your help.
i might try working backwards from your expression.
cheers
May 3rd 2010, 01:04 AM
Kbotz

oh right, my bad!
i can't subtract properly
0-(-s) = s
not -s. lol
May 3rd 2010, 01:07 AM
mr fantastic

Quote:

Originally Posted by Kbotz

i'm a little unsure of how you got s+2 for the y equation, but thanks for your help.
i might try working backwards from your expression.
cheers

Done correctly, $s \times (1) + (s + 1) \times (2)$ will give you y.
May 3rd 2010, 01:56 AM
Kbotz

yeah figured them both out!
and got the final transforms too!
i was on the right track for x(t) from the start!
but cheers again!