Solve the following initial value problem : $$y''(t)+y(t)=f(t)=\begin{cases} e^t & \text{if} \ \ t>0 \\1 & \text{if} \ \ t\leq 0 \end{cases}\quad \text{with} \ y(0)=0,t\in \mathbb{R}$$
I took Fourier transform on both sides and got an expression like $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty y''(t)e^{i\omega t}dt+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty y(t)e^{i\omega t}dt=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0 1.e^{i\omega t}dt+\frac{1}{\sqrt{2\pi}}\int_0^\infty e^t.e^{i\omega t}dt$$$$\implies -\omega^2Y(\omega)+Y(\omega)=\frac{1}{i\omega}+\frac{1}{1+i\omega}\bigg[e^{(1+i\omega)t}\bigg]^\infty_0$$
Now the second integral on the RHS seems to tend to infinity, which is absurd. I sense that I'm doing something wrong as this happens to be a complex integral, so $e^{(1+i\omega)t}$ shouldn't tend to infinity as $\sin t$ and $\cos t$ have their maximum value $1$ over $\mathbb{R}$. What is the possible step that should be taken afterwards in this calculation? Any help is appreciated.