Find unique solution of $y′′ + y = f$ using $y(0) = y′(0) = 0$ and periodic function $f(t) = t$ if $0 \leq t < 2\pi$
Attempted work:
$L[y'' + y ] = L[f(t)]$
$L[y''] + L[y] = L[f(t)]$
$s^2 L[y] -sy(0) -y'(0)+L[y] = L[f(t)]$
$(s^2 + 1) L[y] = L[f(t)]$
I only managed to do that much. I am having trouble with the periodic function and setting up the right hand side of the equation. I am trying to set it up as an integral. I would really appreciate for any help setting up this equation. Thanks in advance!
Suppose $f$ is $T$-periodic. Then $({\cal L} f)(s) = \int_0^\infty f(t) e^{-st} dt = \sum_{n=0}^\infty \int_{nT}^{(n+1)T} f(t) e^{-st} dt$, and using the substitution $\tau = t-nT$ and the fact that $f$ is $T$-periodic, we get $\int_{nT}^{(n+1)T} f(t) e^{-st} dt = \int_0^T f(\tau)e^{-s(\tau + nT)} d \tau = e^{-s nT} \int_0^T f(\tau)e^{-s\tau} d \tau $, and so, using $1+x+x^2+\cdots = {1 \over 1-x}$, we get $({\cal L} f)(s) = {1 \over 1-e^{-sT}}\int_0^T f(\tau)e^{-s\tau} d \tau $ (with $\operatorname{re} s > 0$).
We have $T=2 \pi$, and $f(t) = t$ (for $t \in [0,T)$), so $\int_0^T f(\tau)e^{-s\tau} d \tau = {1 \over s^2}(1-(1+2 \pi s)e^{-2 \pi s})$, and so $({\cal L} f)(s) = {1 \over s^2(1-e^{-2 \pi s})} (1-(1+2 \pi s)e^{-2 \pi s}) $.