Find the general solution of the given second-order differential equation. y'' + 25y = 0

Solution:

Given, the differential equation is y’’ + 25y = 0

We have to find the solution of the equation.

The differential equation can be rewritten as (D² + 25)y = 0

Where, D = d/dx

Auxiliary equation is m² + 25 = 0

m² = -25

Taking square root,

m = ±5i

Since, there are two roots, the general solution is of the form

\(y=Ae^{xk_{1}}+Be^{xk_{2}}\)

Here, k₁ = +5i, k₂ = -5i

So, \(y=Ae^{5ix}+Be^{-5ix}\)

Therefore, the general solution is \(y=Ae^{5ix}+Be^{-5ix}\).

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Find the general solution of the given second-order differential equation. y'' + 25y = 0

Summary:

The general solution of the given second-order differential equation. y'' + 25y = 0 is \(y=Ae^{5ix}+Be^{-5ix}\).