I had discussed that the boundary value problem y"+y=0 with boundary conditions y'(0)=1 and y(pi/2) = 0 yesterday.

There was no solution to this because there would be 2 different values for the arbitrary constant c2.

Now, what change would take place if we had y" +y= x? and the same boundary conditions?

Will it still have no solution ?

The particular solution, in that case, will be of the form y=a+bx ... When I apply the boundary conditions, I see that the a is -pi/2 and b is 1.

This is wrong.
so, does this make the solution be y = x-pi/2? I also see that this merely fails to satisfy the equation given.

I sense that i'm overlooking something...