My exercise is to solve the differential equation
$$(x+1)y''(x)+xy'(x)-y(x)=1$$
With the boundary conditions $y(0)=0$ and $y(1)=0$, and using it's Green's function.
On a physics class, we were solving the $u''(t)+2\beta u(t)+\omega_0^2u(t)=f(t)$ the following way:
The operator is $$L(D)=D^2+2\beta D+\omega_0^2$$ and the green function is $$G(t)=\int_{-\infty}^{\infty}\mathrm{d}\omega \frac{1}{2\pi L(i\omega)}e^{i\omega t}$$ And then the solution is $$u(t)=\int \mathrm{d}\tau f(\tau)G(t-\tau)$$
But now, the $L(D)$ is dependent on $x$ as well: $L(D)=(x+1)D^2+xD-1$, so the Green's function is: $$G(x)=\int_{-\infty}^{\infty}\mathrm{d}\omega \frac{1}{2\pi L(i\omega)}e^{i\omega x}$$ $$G(x)=\int_{-\infty}^{\infty}\mathrm{d}\omega \frac{1}{2\pi} \frac{1}{-(x+1)\omega^2+i\omega x -1}e^{i\omega x}$$
Is it right? Should I just evaluate that integral and go on?
Taking the Fourier transform, you should consider that $$ x^n\rightarrow \left(i\frac{d}{d\omega}\right)^n\qquad \frac{d^n}{dx^n}\rightarrow(-i\omega)^n. $$ So, after Fourier transform, you get again a differential equation given by $$ \left(i\frac{d}{d\omega}+1\right)[(-i\omega)^2{\hat G(\omega)}] +i\frac{d}{d\omega}[-i\omega{\hat G(\omega)}]-{\hat G(\omega)}=1. $$