# Thread: showing integral eq implies bvp

1. ## showing integral eq implies bvp

I'm working through an example in my notes:

Suppose y = y(x) satisfies the integral equation

..............(1)

Show it implies a boundary value problem for y.

Splitting integral at x = t:

..............(2)

The example then says by eliminating the integral term between equations 1 and 2 implies the differential equation y'' - 15y = -9x

I can't see where this has come from at all, could someone please explain? Which integral term??

2. Put: $\displaystyle A = \int_0^1 {e^{3\left| {x - t} \right|} y\left( t \right)dt}$ then: (1) $\displaystyle y-x=A$ and (2) $\displaystyle y'' = 9A + 6y$ Now we replace A of (1) in (2) $\displaystyle y'' = 9\left( {y - x} \right) + 6y \Leftrightarrow y'' - 15y = - 9x$

3. Originally Posted by hunkydory19
I'm working through an example in my notes:

Suppose y = y(x) satisfies the integral equation

..............(1)

Show it implies a boundary value problem for y.

Splitting integral at x = t:

..............(2)

The example then says by eliminating the integral term between equations 1 and 2 implies the differential equation y'' - 15y = -9x

I can't see where this has come from at all, could someone please explain? Which integral term??

The integral term is $\displaystyle \int_0^1 e^{3|x-t|}dt$, of course, the only integral that appears in those two equations! Multiply the first equation by 9 and subtract from the second equation.