Answer:
[tex]y(t)=\frac{1}{4} tsin(2t)+\frac{3}{2} sin(2t)+5cos(2t)[/tex]
Step-by-step explanation:
we are given
y(0)=5
y'(0)=3
[tex]y''+4y=cos(2t)[/tex]
we can take Laplace transform both sides
[tex]s^2Y(s)-sy(0)-y'(0)+4Y(s)=\frac{s}{s^2+4}[/tex]
now, we can plug values
[tex]s^2Y(s)-5s-3+4Y(s)=\frac{s}{s^2+4}[/tex]
now, we can solve for Y(s)
[tex]Y(s)=\frac{s}{(s^2+4)^2}+\frac{5s+3}{(s^2+4)}[/tex]
now, we can take inverse Laplace transform both sides
we can use Laplace transform table
and we get
[tex]y(t)=\frac{1}{4} tsin(2t)+\frac{3}{2} sin(2t)+5cos(2t)[/tex]
