Question

Use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed. y'' + 4y = cos 2t, y(0) = 5, y'(0) = 3

Answer 1

Answer:

$y(t)=\frac{1}{4} tsin(2t)+\frac{3}{2} sin(2t)+5cos(2t)$

Step-by-step explanation:

we are given

y(0)=5

y'(0)=3

$y''+4y=cos(2t)$

we can take Laplace transform both sides

$s^2Y(s)-sy(0)-y'(0)+4Y(s)=\frac{s}{s^2+4}$

now, we can plug values

$s^2Y(s)-5s-3+4Y(s)=\frac{s}{s^2+4}$

now, we can solve for Y(s)

$Y(s)=\frac{s}{(s^2+4)^2}+\frac{5s+3}{(s^2+4)}$

now, we can take inverse Laplace transform both sides

we can use Laplace transform table

and we get

$y(t)=\frac{1}{4} tsin(2t)+\frac{3}{2} sin(2t)+5cos(2t)$