How does the following differential equation imply the integral equation? The original differential equation:
$\begin{cases}
y''+\left(1+t^2 \right )y=0, &t>0 \\ 
 y\left ( 0 \right )= 1, y'\left ( 0 \right )=0
\end{cases}$
The corresponding integral equation: $y\left ( t \right ) = \cos \left ( t \right ) + \int_{0}^{t}\sin \left ( \tau -t \right )\tau^2 y\left ( \tau \right )d\tau$.
I have already shown that the soltions to the latter one are indeed the solutions to the former one. What is left is just the converse. A method figrued out is to use the uniqueness and the existence od the solution to the former eq.: since the latter one has solutions to the former one, and if we can show that the latter one indeed has a solution (has been done), then by the uni. and exis. we may draw the conlusion that two eq.'s have the same solution.
However, I seek a more direct way to show that the diff. eq. implies the int. eq..
 A: Alright my frd solved it. Not difficult.
The original diff. eq. is equivalent to 
$\tau^2 y\left ( \tau \right )= -y\left ( \tau \right )-y''\left ( \tau \right )$. 
Multiply both sides by the sine function and then integrate from $0$ to $t$:
$\int_{0}^{t}\sin\left ( \tau-t \right )\tau^2 y\left ( \tau \right ) d\tau= -\int_{0}^{t}\sin\left ( \tau-t \right )y\left ( \tau \right )d\tau-\int_{0}^{t}\sin\left ( \tau-t \right )y''\left ( \tau \right )d\tau$
Apply integration by parts on the R.H.S.:
$\int_{0}^{t}\sin\left ( \tau-t \right )\tau^2 y\left ( \tau \right ) d\tau = \int_{0}^{t} y\left ( \tau \right )d\cos\left ( \tau-t \right )-\int_{0}^{t} \sin\left ( \tau-t \right ) d y'\left ( \tau \right )=y\left ( t \right )- y\left ( 0\right )\cos t - \int_{0}^{t} y'\left ( \tau \right )\cos\left ( \tau-t \right ) d \tau - 0+\left ( - y'\left ( 0 \right )\sin\left ( t \right )\right )+\int_{0}^{t} y'\left ( \tau \right )\cos\left ( \tau-t \right ) d \tau$
Plug in the initial values in the R.H.S.:
$\int_{0}^{t}\sin\left ( \tau-t \right )\tau^2 y\left ( \tau \right ) d\tau = y\left ( t \right )-\cos t$
which is what we desired.
