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
$\displaystyle
sX + 2sY + X = 1/s^2$ 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 $ eqn (1')
Multiply equation 2 by s to remove the 's' on the RHS.
$\displaystyle
-s^2x - s^2Y + sY = 1$ 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
then when s =1, substitute in A =2 to find out B
$\displaystyle s = 1, B = -3$
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.