Solving differential equations

[Mildred]

Hi

I am not very advanced in calculas.

I need to solve for X(t)/Y(t) when t -> infinity.

X'(t)+aX(t)+cX(t)-bY(t)=0 ....1 X(0)=0
Y'(t)+bY(t)+cX(t)=0 .....2 Y(0)=0

So I thought of deriving both equations to get:

X''(t)+aX'(t)+cX'(t)-bY'(t)=0 ...3
Y''(t)+bY'(t)+cX'(t)=0 ...4

Then substituting 2 into 3 & substituting 1 into 4

X''(t)+aX'(t)+cX'(t)-b*[bY(t)]-bcX(t)=0 ....5 Y''(t)+bY'(t)+cX(t)[a+c]+bY(t)=0 .....6

Then substituting 1 into 5 Then substituting 2 into 6

X''(t)+aX'(t)+cX'(t)+bX'(t)+baX(t)=0 Y''(t)+[a+b+c]Y'(t)+[ab+cb+b]Y(t)=0

Taking the Laplace transform Taking the Laplace transform

X(s)[s^2+s+a+c+sa+sc+sb+ba]=0 Y(s)[s^2+sa+2sb+sc+sab+scb-s-b]

Then

X(s)/Y(s)=[s^2+sa+2sb+sc+sab+scb-s-b]/[s^2+s+a+c+sa+sc+sb+ba]

This is where I get stuck as I don't know how to transform it back to get X(t)/Y(t)

Any help would be much appreciated.

Thank you

 

[JohnZ]

X''(t)+aX'(t)+cX'(t)+bX'(t)+baX(t)=0 Y''(t)+[a+b+c]Y'(t)+[ab+cb+b]Y(t)=0

you don't need Laplace transform. instead you can use auxiliary equation to find X(t) and Y(t).