I believe this to be correct but i'd really appreciate a quick glance over to make sure my workings correct. anyway here we go
Express the following ODE's solution as an integral $$ f''(t) + \lambda^{2} f(t) = g(t),~\text{With } f(0)=1,~f'(0)=0$$
Applying LT transformations to both sides give:
$$\mathcal{L}[f''(t)]+\lambda^{2}\mathcal{L}[f(t)]=\mathcal{g(t)},~\text{With }\mathcal{L}[f(0)]=\mathcal{L}[1],\mathcal{L}[f'(0)]=\mathcal{L}[0]$$ which breaks into the following \begin{align*} \mathcal{L}[f''(t)] &= s^{2}\bar{f}(s)-sf(0)-f'(0) = s^{2}\bar{f}(s)-s\\ \mathcal{L}[f(t)] &= \bar{f}(s) \\ \mathcal{L}[g(t)] &= \bar{g}(s) \\ \mathcal{L}[f(0)] &= \bar{f}(0) = \mathcal{L}[1] = \frac{1}{s} \\ \mathcal{L}[f'(0)] &= s\bar{f}(0)-f(0) = 1-1= 0 \\ \\ \text{Combining all this we get} \\ \\ s^2\bar{f}(s)-s+\lambda^{2}\bar{f}(s)&=\bar{g}(s), \text{With } \bar{f}(0)= \frac{1}{s} \implies \\ \bar{f}(s)(s^2+\lambda^2) &= \bar{g}(s)+s \implies\\ \bar{f}(s) &= \frac{\bar{g}(s) + s}{s^{2}+\lambda^2} = \frac{\bar{g}(s)}{s^{2}+\lambda^2} + \frac{ s}{s^{2}+\lambda^2}\\ \end{align*} From here we consider the two equations separately and apply the inverse laplace transformation $$\mathcal{L}^{-1}\left[\frac{\bar{g}(s)}{s^{2}+\lambda^2}\right] = \mathcal{L}^{-1}\left[\frac{\bar{g}(s)}{\lambda}\frac{\lambda}{s^{2}+\lambda^2}\right] = \frac{1}{\lambda}\mathcal{L}^{-1}\left[\bar{g}(s)\frac{\lambda}{s^{2}+\lambda^2}\right]=\frac{1}{\lambda}\mathcal{L}^{-1}\left[\mathcal{L}[g(t)]\mathcal{L}\left[\sin{\lambda t}\right]\right]$$ applying the convolution rule we get $$\mathcal{L}^{-1}\left[\frac{\bar{g}(s)}{s^{2}+\lambda^2}\right] =\frac{1}{\lambda} \int_{0}^{t}g(s-\tau)\sin{\lambda \tau}~d\tau$$ next we have $$\mathcal{L}^{-1}\left[\frac{s}{s^{2}+\lambda^2}\right] =\cos{\lambda t}$$ combining we have $$f(t) = \cos{\lambda t} + \frac{1}{\lambda} \int_{0}^{t}g(t-\tau)\sin{\lambda \tau}~d\tau$$ as the solution.
Any help would be much appreciated, and thank you for taking the time to read over this.