I need help understanding a part of this question.
Let $a.) y''+4y = \delta (x)$, $y(0)=y'(0)=0$.
and $b.) y'' + 4y = f(x)$, $y(0)=y'(0)=0$
where $f(x)$ is some continuous function of finite exponential order. Let $y_{\delta}(x)$ and $y_{f}(x)$ denote solutions of the initial value problems $a.$ and $b.$
So the first part is to solve for $y_{\delta}(x)$ which I got $(1/2) sin(2x)$
However, I dont know how to show that the Laplace transform of $y_{f}(x)$ is given by $y\hat f(s) = y\hat \delta(s)f \hat (s)$, where $y\hat f(s)$ and $f\hat (s)$ denote the Laplace transforms of $f(x)$ and $y_{\delta} (x)$. Use convolution to deduce that
$y_{f}(x) = \int_0^x y_{\delta}(x-t) f(t) \mathrm{d}t$
and then find $y_{f} (x) when f(x)=2.
Any help would be great. A little lost on the second part.
Note that from (a) you have
$$Y_{\delta}(s)=\frac{1}{s^2+4}\tag{1}$$
and from (b) you have
$$Y_{f}(s)=\frac{F(s)}{s^2+4}\tag{2}$$
Comparing (1) and (2) you get
$$Y_{f}(s)=Y_{\delta}(s)F(s)\tag{3}$$
This relation means that the system's response to any function is completely determined by its impulse response $y_{\delta}(t)$. Since multiplication of the Laplace transforms corresponds to convolution of the original functions, (3) implies
$$y_f(x)=\int_0^xy_{\delta}(x-t)f(t)dt=\int_0^xf(x-t)y_{\delta}(t)$$
With $f(x)=2$ and $y_{\delta}=\frac12\sin 2x$ you have
$$y_f(x)=\int_0^x\sin(2t)dt=\sin^2x$$