Question

Find the general soltuion of the fourth-order ODE y''''-2y''+y=0

Answer 1

Use a trial function (sometimes called Ansatz) of y(x) = e^(ax).

Substitute this into your equation and see what you get.

Answer 2

Do you know about the characteristic equation?

Here it is m^4 -2m^2 + 1 = 0
It has 4 roots, which we can find by substituting n= m^2:
n^2 -2n+1 = 0;
n= +/- 1
and m= sqrt(n), so m= 1, -1, i, -i
the solution to the ODE is c e^(m t), where c is a constant, m is a root of the char eq.

Answer 3

For the record, using the Ansatz above is how you arrive at the characteristic equation.

Answer 4

oops, it's n=1 (repeated root), and m= +1 or -1

Answer 5

for a repeated root, the anwer is c1 e^mt +c2 t e^mt

Answer 6

\[y= c_1e^{t}+c_2te^{t}+c_3e^{-t}+c_4te^{-t}\]