Find the general solution of the given second-order differential equation. 2y'' + y' = 0

Solution:

Given, the differential equation is 2y’’ + y = 0

We have to find the solution of the equation.

The differential equation can be rewritten as (2D² + 1)y = 0

Where, D = d/dx

Auxiliary equation is 2m² + 1 = 0

m² = -1/2

Taking square root,

m = ±(1/2)i

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

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

Here, k₁ = +(1/2)i, k₂ = -(1/2)i

So, \(y=Ae^{\frac{1}{2}ix}+Be^{-\frac{1}{2}ix}\)

Therefore, the general solution is \(y=Ae^{\frac{1}{2}ix}+Be^{-\frac{1}{2}ix}\).

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Find the general solution of the given second-order differential equation. 2y'' + y' = 0

Summary:

The general solution of the given second-order differential equation. 2y'' + y = 0 is \(y=Ae^{\frac{1}{2}ix}+Be^{-\frac{1}{2}ix}\).