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

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- Apr 28th 2010, 01:19 AMKbotzSystem 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 - Apr 28th 2010, 05:02 AMmr fantastic
- Apr 28th 2010, 10:23 AMKbotz
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. - Apr 28th 2010, 06:39 PMmr fantastic
- Apr 28th 2010, 08:13 PMKbotz
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 PMKbotz
hey,

i ran into more trouble doing this problem.

i have no idea whether my answer is right.

**transformed equations as in earlier post**$\displaystyle

sX + 2sY + X = 1/s^2$ eqn (1)

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

**Multiply equation 1 by $\displaystyle s^2$ to remove the $\displaystyle s^2$ on the RHS.**$\displaystyle s^3X + 2s^3Y + s^2X = 1 $

$\displaystyle

$**eqn (1')**$\displaystyle -s^2x - s^2Y + sY = 1$

Multiply equation 2 by s to remove the 's' on the RHS.

$\displaystyle

$**eqn (2')**

equation 1' & 2' commonly factorized in terms of X

$\displaystyle

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

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

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

$\displaystyle X(s^3 + s^2) + s^3Y - sY = -s-1$.....**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:**

$\displaystyle s^3Y + sY = 2-s

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

**Make Y the subject**

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

**Now using partial fractions**$\displaystyle 2-s = A/s + B/(s^2+1)$

cross multiply

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

**when****s = 0, B=0 and A = 2**$\displaystyle s = 1, B = -3$

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

**Taking the inverse laplace transform of the partial fraction:**

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

**$\displaystyle y(t) = 2 - 3sin(t)$**

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 AMmr fantastic
I get:

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

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

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 AMKbotz
- May 3rd 2010, 01:04 AMKbotz
oh right, my bad!

i can't subtract properly

0-(-s) = s

not -s. lol - May 3rd 2010, 01:07 AMmr fantastic
- May 3rd 2010, 01:56 AMKbotz
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!