# Find the general solution of the given second-order differential equation. y'' + 4y' + 4y = 0

**Solution:**

Given, the differential equation is y’’ + 4y’ + 4y = 0

We have to find the solution of the equation.

The differential equation can be rewritten as (D^{2} + 4D + 4)y = 0

Where, D = d/dx

Auxiliary equation is m^{2 }+ 4m + 4 = 0

Using quadratic formula,

\(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here, a = 1, b = 4, c = 4

\(\\x=\frac{-4\pm \sqrt{(4)^{2}-4(1)(4)}}{2(1)}\\x=\frac{-4\pm \sqrt{(16-16}}{2}\\x=\frac{-4}{2}\)

x = -2

Since, there is only one root, the general solution is of the form

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

Her, k_{1} = -2

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

Taking out common term,

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

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

## Find the general solution of the given second-order differential equation. y'' + 4y' + 4y = 0

**Summary:**

The general solution of the given second-order differential equation. y'' + 4y' + 4y = 0 is \(y(x)=(C_{1}+C_{2}x)e^{-2x}\).

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