Laplace transforms of differential equations

[TyErd]

Homework Statement

I've attached the problem.

Homework Equations

L(1) = 1 /s
L(t^n) = n!/s^(n+1)

The Attempt at a Solution

because the question only asks for X(s) I only considered the x' + x + 4y=3 equation.
I applied laplace tranforms and got:

s X(s) - x(0) + 1/s^2 + 4/s^2 = 3/s. since x(0) = 0 it gets canceled out so..
s X(s) + 5/s^2 = 3/s
s X(s) = 3/s - 5/s^2

therfore
X(s) = (3/s - 5/s^2) / s

only problem is, i didn't use the other equation which makes me feel I've done it wrong.

 

[RoshanBBQ]

In

sX(s) - x(0) + \frac{1}{s^2} + \frac{4}{s^2} = \frac{3}{s}
how are you getting
\frac{1}{s^2} + \frac{4}{s^2}
?

 

[LCKurtz]

Homework Statement

I've attached the problem.

Homework Equations

L(1) = 1 /s
L(t^n) = n!/s^(n+1)

The Attempt at a Solution

because the question only asks for X(s) I only considered the x' + x + 4y=3 equation.
I applied laplace tranforms and got:

s X(s) - x(0) + 1/s^2 + 4/s^2 = 3/s. since x(0) = 0 it gets canceled out so..
s X(s) + 5/s^2 = 3/s
s X(s) = 3/s - 5/s^2

therfore
X(s) = (3/s - 5/s^2) / s

only problem is, i didn't use the other equation which makes me feel I've done it wrong.

It's well you should feel you've done it wrong.

You have two equations in two unknowns, x and y. When you transform, each equation will have an X(s) and Y(s). You left out the Y(s). You will get two equations in the unknowns X(s) and Y(s) which you can solve for X(s).

 

[TyErd]

Omg I feel so stupid right now, i can't believe i was trying to laplace transform a x and y variable >.<

so it should be s X(s) - x(0) + X(s) + 4 Y(s) = 3/s and s Y(s) - y(0) + 9 X(s) + Y(s) = 0 simplify and it becomes

s X(s) + X(s) + 4 Y(s) = 3/s and s Y(s)+ 9 X(s) + Y(s) = 0

 

[TyErd]

rearrange the second equation and it gets me Y(s) = -9 X(s) / (s+1)

s X(s) + X(s) - 36 X(s) / s+1 = 3/s
X(s) [s+1- 36/(s+1) ] = 3/s
X(s) = 3(s+1) / x(s^2 + 2s - 35)

not sure if calculations are correct.