Green's Function Ode 1 Can any one help me to how to find a) Green's function for both of them and how can i solve these by using Greens function
1)$\frac{d}{dt}(\frac{1}{t+1}\frac{dy}{dt})=f(t); \quad   y(0)=y(1)=0$
2)$\frac{d}{dt}((t+1)\frac{dy}{dt})=f(t); \quad   y(0)=1;y(1)=-1$
 A: Start with the solutions of the homogeneous equation:
$$
        \frac{d}{dt}\left(\frac{1}{t+1}\frac{dy}{dt}\right) = 0.
$$
This system is integrable
$$
            \frac{1}{t+1}\frac{dy}{dt} = C \\
               \frac{dy}{dt} = C(t+1)
$$
The general solution involves two constants $D$ and $E$:
$$
                 y = D(t+1)^{2}+E.
$$
The solutions satisfying $y(0)=0$ have the form $D(t^{2}+2t)$. The solutions satisfying $y(1)=0$ have the form $E(t^{2}+2t-3)$. The Green function $y_{x}(t)$ is obtained by piecing together the two solutions so that they (a) meet continuously at $x \in [0,1]$ and so that (b) $\frac{1}{t+1}y'(t)$ has a unit jump at $t=x$, which intuitively gives
$$
          \frac{d}{dt}\frac{1}{t+1}\frac{d y_{x}(t)}{dt} = \delta_{x}(t).
$$
Continuity of $y_{x}(t)$ at $t=x$ with $y_{x}(0)=0$ and $y_{x}(1)=0$ forces there to be a constant $C$ such that
$$
                y_{x}(t) = C\left\{\frac{t^{2}+2t}{x^{2}+2x}\chi_{[0,x]}(t)+\frac{t^{2}+2t-3}{x^{2}+2x-3}\chi_{[x,1]}(t)\right\}
$$
Then, for $t \ne x$,
$$
         \frac{1}{t+1}y_{x}'(t) = C\left\{ \frac{2}{x^{2}+2x}\chi_{[0,x]}(t)+
                       \frac{2}{x^{2}+2x-3}\chi_{[x,1]}(t)\right\}
$$
The jump in this function at $t=x$ is
$$
\begin{align}
  y_{x}(x+0)-y_{x}(x-0)
     & = 2C\left\{\frac{1}{x^{2}+2x-3}-\frac{1}{x^{2}+2x}\right\} \\
     & =\frac{6C}{(x^{2}+2x-3)(x^{2}+2x)}.
\end{align}
$$
Requiring the jump to be $1$ gives
$$
               C = \frac{1}{6}(x^{2}+2x-3)(x^{2}+2x).
$$
So the Green function is
$$
\begin{align}
   y_{x}(t) & = \frac{1}{6}(x^{2}+2x-3)(t^{2}+2t)\chi_{[0,x]}(t)+ \\
            & + \frac{1}{6}(x^{2}+2x)(t^{2}+2t-3)\chi_{[x,1]}(t).
\end{align}
$$
The solution of (1) is
$$
\begin{align}
   y(t) & = \int_{0}^{1}y_{x}(t)f(x)dx \\
     & = \frac{1}{6}(t^{2}+2t-3)\int_{0}^{t}(x^{2}+2x)f(x)dx+ \\
     & + \frac{1}{6}(t^{2}+2t)\int_{t}^{1}(x^{2}+2x-3)f(x)dx.
\end{align}
$$
Check Solution: Directly verify that $y(0)=y(1)=0$. Then
$$
\begin{align}
   y'(t) & = \frac{1}{6}(2t+2)\int_{0}^{t}(x^{2}+2x)f(x)dx+ \\
         & + \frac{1}{6}(2t+2)\int_{t}^{1}(x^{2}+2x-3)f(x)dx. \\
   \frac{1}{t+1}y'(t) & = \frac{1}{3}\int_{0}^{t}(x^{2}+2x)f(x)dx+ \\
         & + \frac{1}{3}\int_{t}^{1}(x^{2}+2x-3)f(x)dx. \\
   (\frac{1}{t+1}y')' & = \frac{1}{3}(t^{2}+2t)f(t)-\frac{1}{3}(t^{2}+2t-3)f(t) \\
                      & = \frac{1}{3}3f(t) = f(t).
\end{align}
$$
For (2) choose $0$ endpoint conditions and, then, when finished add a linear combination of the solutions of the homogeneous differential equation in order to match the non-zero endpoint conditions.
