How do you find the transfer function of a system described by two simultaneous de's, ie,

-(D+1)y1(t) + (2D + 1)y2(t) = x(t)

-(D+2)y1(t) + (D+1)y2(t) = 0

Those are negative signs.

x(t) is given.

I want to find the impulse response "relating outputs y1(t) and y2(t) to the input x(t)"

im just wondering whether there will be two different transfer functions and thus two different impulse responses...

 

How do you find the transfer function of a system described by two simultaneous de's, ie,

-(D+1)y1(t) + (2D + 1)y2(t) = x(t)

-(D+2)y1(t) + (D+1)y2(t) = 0

Those are negative signs.

x(t) is given.

I want to find the impulse response "relating outputs y1(t) and y2(t) to the input x(t)"

im just wondering whether there will be two different transfer functions and thus two different impulse responses...

You can treat the operators (e.g., (D+1) and so on) as if they were coefficients in a pair of equations in two unknowns, and solve just as you would a system of two linear equations in two unknowns.

Multiply the top equation by -(D+2) and the bottom equation by (D+1). You're actually not multiplying, but applying these operators to each equation.

The preceding operations yield:
(D+2)(D+1)y1 - (2D+1)(D+2)y2 = -(D+2)x
-(D+2)(D+1)y1 +(D+1)(D+1)y2 = 0
(I have omitted the t parameter from y1, y2 and x.)

Adding the first equation to the second eliminates y1, and results in:
-(2D+1)(D+2)y2 + (D+1)(D+1)y2 = -(D+2)x

Simplifying, we get:
(-2D^2 - 5D -2 + D^2 + 2D + 1)y2 = -(D+2)x

and then,
(-D^2 -3D -1)y2 = -(D+2)x

or, finally,
(D^2 + 3D + 1)y2 = (D+2)x

The last equation is second order in y2, and as you said, x is known. Presumably you know how to solve second-order DEs.

After you get y2, you can substitute it into your second equation to find y1.

Is that enough to get you going?
Mark