Question

y"-4y'+9y=t,y(0)=0,y'(0)=1, find the transfer function

Answer 1

Laplace transform of $y''-4y'+9y=t$, is given as

$$L( y''-4y'+9y)=L(t)$$

$$L(y'')=s^2L(y)-sy(0)-y'(0)$$

$$L(y')=sL(y)-y(0)$$

$$L(t)=\dfrac{1}{s^2}$$

Putting the above values in the differential equations, we get,

$$s^2L(y)-1-4sL(y)+9L(y)=\dfrac{1}{s^2}$$ $$L(s) (s^2-4s+9)=\frac{1}{s^2}+1$$

or, 

$$L(y)=\frac{1+s^2}{s^2(s^2-4s+9)}$$