# Math Help - System of DE's: Laplace transfomation

1. ## 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

2. 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?

3. 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.

Y = 1+s/s^3

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

Thanks Fantastic.

4. 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

5. 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!

6. 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.

7. 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.

8. 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

i can't subtract properly
0-(-s) = s
not -s. lol

10. 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.

11. 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!