y=a1e^x+a2e^2x+a3e^3x then obtain diffrential equation?

Answer 1

The roots of the characteristic equation are 1, 2, 3.

The characteristic equation is
(r - 1)(r - 2)(r - 3) = (r² - 3r + 2)(r - 3) = r³ - 6r² + 11r - 6

There's no "particular solution" so the original equation was homogeneous.

So the original equation must have been y''' - 6y'' + 11y' - 6y = 0.

I assume you don't need to derive the boundary conditions based on the coefficients given.

Disclaimer: it's been 13 years since I took diff eq.

Answer 2

Easy, the powers of the exponentials tell you the roots of the auxiliary equation. The roots are basically
m=1,2,3 so the corresponding polynomial (one of them) is
(m-1)(m-2)(m-3)=0
m^3-6m^2+11m-6=0

Replace the m's with the y's and your differential equation is
y'''-6y''+11y'-6y=0