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### Second-Order Linear Homogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 8y &= 0 \\
y(0) &= 8 \\
y'(0) &= -10
\end{align}
$$

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©2010, 2014 Andrew G. Bennett

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This is a second-order linear constant-coefficient initial value problem. First we find the general solution.

*Step 1: Write the equation in operator form.*
$$
(D^2 - 6D + 8)y = 0
$$
*Step 2: Find the roots.*
The equation factors into $(D - 4)(D - 2)y = 0.$
So our roots are 4 and 2.

*Step 3: Write the general solution.*

$$ y(x) = c_1\exp(4x) + c_2\exp(2x) $$ Now that we have the general solution, we plug in our initial values to find the solution to the initial value problem. First we compute, $ y'(x) = 4c_1\exp(4x) + 2c_2\exp(2x).$ Then we plug in $ x=0 $ to get the following equations. $$ \begin{align} y(0) = c_1 + c_2 &= 8 \\ y'(0) = 4c_1 + 2c_2 &= -10 \end{align} $$ We solve these equations to get $ c_1= -13 $ and $ c_2 = 21.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.

$$
y(x) = -13\exp(4x) + 21\exp(2x)
$$
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©2010, 2014 Andrew G. Bennett