# Find the general solution of the given higher-order differential equation. y(4) - 2y'' + y = 0

**Solution:**

The auxiliary equation of the given homogenous differential equation can be written as:

m^{4} - 2m^{2 }+ 1 = 0

m^{4} - m^{2} - m^{2 }+ 1 = 0

m^{2}(m^{2} - 1) - 1(m^{2 }- 1) = 0

(m^{2} - 1)(m^{2} - 1) = 0

(m - 1)(m + 1)(m - 1)(m + 1) =0

The roots of the equation are:

m = 1, 1, -1, -1

Since roots are repeating in pairs we can write the general solution as sum of two parts. The first part being y_{1} and y_{2}.

y_{1} = (c_{1} + c_{2}x)\(e^{x}\)

y_{2} = (c_{3 }+ c_{4}x)\(e^{-x}\)

And

y = y_{1} + y_{2} = (c_{1} + c_{2}x)\(e^{x}\) + (c_{3}+ c_{4}x)\(e^{-x}\)

## Find the general solution of the given higher-order differential equation. y(4) - 2y'' + y = 0

**Summary:**

The general solution of the given higher-order differential equation. y(4) - 2y'' + y = 0 is given by

y = y1 + y2 = (c1 + c2x)\(e^{x}\) + (c3+ c4x)\(e^{-x}\)