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:

m⁴ - 2m² + 1 = 0

m⁴ - m² - m² + 1 = 0

m²(m² - 1) - 1(m² - 1) = 0

(m² - 1)(m² - 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 y₁ and y₂.

y₁ = (c₁ + c₂x)\(e^{x}\)

y₂ = (c₃ + c₄x)\(e^{-x}\)

And

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

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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)\(e^{-x}\)