Solving the IVP $y''+4y=e^t, y(0) = 0 , y'(0) = 0 $ by using laplace transfom

Question

I've taken the laplace transform on both sides giving me:

$$s^2L(y)+4L(y) = L(e^t)$$

$$L(y)=1/(s^2+4)*L(e^t)$$

but now I'm a bit stuck.

Answer 1

Hint: The Laplace transform is given as

$$L(y) = \frac{1}{2}L\left(\sin 2t\right)L(e^t)$$

Using the convolution property (I will not explicitly write down the Heaviside step function):

$$y = \frac{1}{2}\sin 2t * e^t=\frac{1}{2}\int_{0}^t\sin 2\tau \cdot e^{t-\tau}d\tau=\frac{1}{2}e^{t}\int_{0}^{t}\sin 2\tau \cdot e^{-\tau}d \tau$$

You can solve the last integral by using integration by parts or by complexifying the integrand to $e^{2i\tau}e^{-\tau}$ (solving this should be easy) the integral is then given as the imaginary part of $$\int_{0}^{t}e^{2i\tau}e^{-\tau}d\tau.$$