Use Laplace to solve differential equation y4ycos3t y02 y01

Use Laplace to solve differential equation y\"-4y=cos3t, y(0)=2, y\'(0)=1?

L{y\"}-4L{y}=L{cos3t}

L{y}(s^2+4)=(s/(s^2+9))+2s+1

L{y}=(s/(s^2+9)(s^2-4))+(2s+1/(S^2-4))

I\'m stuck here... please help!!

Solution

L{y\'\'}-4L(y)=Lcos3t

s^2L{s}-sy(0)-y\'0-4L(y)=s/s^2+9

Where y(0)=2&y\'(0)=1

L(s){s^2-4}-{2s+1}=s/s^2+9

L(s)=s+2s^3+18s+s^2+9/{(s^2+9)(s^2-4)}

=s+s^2(2s+1)+9(2s+1)/{(s^2+9)(s^2-4)}

=s+(s^2+9)(2s+1)/{(s^2+9)(s^2-4)}

=s/{(s^2+9)*(s^2-4)}+(2s+1)/s^2-4)

So,

Y=L^-1{s/{(s^2+9)(s^2-4)}}+L^-1{2s/(s^2-4)}+L^-1{1/(s^2-4)}

Y=L^-1{s/(s^2+9)}*L^-1{(1/(s^2-4))}+L^-1{2s/(s^2-4)}+L^-1{1/(s^2-4)}

Y={cos3t×sinh 2t}/2+2cosh2t+(sinh2t)/2