Solve the following system of ODEs using Laplace transforms

Solve the following system of ODEs using Laplace transforms { x\' + x - 3y = 0 y\' - 4x + 2y = 0 { x (0) = 7 y (0) =0

Solution

x\' = - x + 3y

y\' = 4x - 2y

We begin by applying the Laplace tranform to both sides of each equation:

L(x\') = - L( x ) + L( y )

L(y\') = 4 L( x ) - 2L( y )

Setting X(s) = L(x(t)) and Y (s) = L(y(t)) and using the result that L(f 0 )(s) = sL(f)(s) f(0),

we find that the differential system is transformed into a pair of simultaneous linear algebraic equations for the functions X and Y :

sX(s) 1 = - X(s) + 4Y( s )   

sY (s) = 4 X(s) - 2Y(s)   

which gives,

sX(s) + X(s) - 4 y(s) = 1...................(i)

sY(s) +2 Y(s) - 4 X(s) = 0 ...................(ii)

solving (i) & (ii) for X(s) & Y(s),we get-

X(s) = 4 ( s + 2 ) / {(s-2)(s+5)}

Y(s) = 4 /{(s -2 )(s + 5 )}

We now wish to find the inverse transform of the functions X and Y which will then be solutions of the system. To do this, we use a table of Laplace transforms after decomposing the expressions into sums of simple fractions. Using the method of partial fraction expansions leads to

X(s) = 16 / { 7 (s-2)} + 12 / {7(s+5)}

Y(s) = 4/ {7(s-2)}- 4/ {7(s+5)}

which means,

X(s) =(16/7) / (s-2) + (12/7) / (s+5)

x(t=(16/7)e^2t+ (12/7) e^{-5t .................(III)}

y(t) = (4/7)e^2t -(4/7)e^{-5t ..................(iv)}