Answer to Question #117544 in Complex Analysis for Chandana

Question

Find the Laplace transform of y"-4y'+9y=t ,y(0)=0,y'(0)=1

Answer 1

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

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

L(y'')=$s^2F(s)-sf(0)-f'(0)$

It is given in the question that, y(0)=0 and y'(0)=1.

L(y)=$Y(s)$

Hence, L(y'') = $s^2Y(s)-s.0-1=s^2Y(s)-1.$

Similarly, L(y')=$sY(s)-y(0)=sY(s)$

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

Putting the above values in the differential equations,

we get,

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

or, $Y(s) (s^2-4s+9)=\frac{1}{s^2}+1$

or, $Y(s)=\frac{1+s^2}{s^2(s^2-4s+9)}$ (Answer)