Differiential equation?

Question

Why for diff equation is e^(alpha*x) substituted say for

d2y/dx2 - ky(x) = 0
say K = 4
d2y/dx2 - 4y(x) = 0

y(x) = e^(alpha*x)

substituting e^(alpha*x) we find
(alpha^2 -4)y(x) = 0
alpha = +/- 2

How does this work?

Answer 1

The general solution is in fact y = Ae^(αx) + Be^(-αx)

y = e^(αx) is only a specific solution with A = 1 and B = 0

Check

y' = A. αe^(αx) + B. -αe^(-αx)
= α(Ae^(αx) - Be^(-αx))

y'' = α(A. αe^(αx) - B. -αe^(-αx))
= α²(Ae^(αx) + Be^(-αx))
= α²y

ie y'' - α²y = 0

You can take this further with α² <0 ie y'' + β²y = 0
Then the solution is
y = Ae^(iβx) + Be^(-iβx)
which can be reduced to y = Ccos(βx + ε) + Dsin(βx + ε)) the general SHM solution

Answer 2

It's a consequence of the fact that d/dx[e^(alpha*x)] = alpha*e^(alpha*x), which implies that d^n/dx^n[e^(alpha*x)] = alpha^n*e^(alpha*x).

Because differentiation is a linear operation, it follows that for any polynomial p, p(d/dx)e^(alpha*x) = p(alpha)e^(alpha*x).

Since a homogeneous linear ordinary differential equation with constant coefficients has the form p(d/dx)y(x) = 0, for some polynomial p, substituting y(x) = e^(alpha*x) yields the equation p(alpha)e^(alpha*x) = 0, and since e^(alpha*x) is never zero, it follows that the equation is satisfied if and only if p(alpha) = 0.

Answer 3

The de you have is of the class of order 2 with constant co-efficients.

It will have the solution of y = ae^rx + be^-rx where

r^2 - k=0 which leads to r=+/- k^0.5

Answer 4

it is called complementary function. differential equation, in general have infinte series as its solution. to find its roots, we need to find its roots which is called complementary function. if right hand side is zero, particular integral vanishes and so the solution is only complimentary function for that differential equation.