Find the general solution of the given second-order differential equation. 8y'' + y' = 0

Solution:

Given, the differential equation is 8y’’ + y’ = 0

We have to find the general solution.

Now, 8x² + x = 0

x(8x + 1) = 0

x₁ = 0

8x + 1 = 0

8x = -1

x₂ = -1/8 

The roots of the differential equation are distinct and real.

So, the general solution is of the form,

\(y = C_{1}e^{m_{1}x}+C_{2}e^{m_{2}x}\)

Here, m₁ = 0 and m₂ = -1/8 

\(y = C_{1}e^{0x}+C_{2}e^{(-1/8)x}\)

We know, e⁰ = 1

So, \(y = C_{1}(1)+C_{2}e^{(-1/8)x}\)

Therefore, the general solution is \(y = C_{1}+C_{2}e^{(-1/8)x}\)

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Find the general solution of the given second-order differential equation. 8y'' + y' = 0

Summary:

The general solution of the given second order differential equation 8y’’ + y’ = 0 is \(y = C_{1}+C_{2}e^{(-1/8)x}\)