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### Second-Order Linear Homogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} + 4\frac{dy}{dx} - 32y &= 0 \\
y(0) &= -8 \\
y'(0) &= 6
\end{align}
$$

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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.*
$$
(D^2 + 4D - 32)y = 0
$$
*Step 2: Find the roots.*
The equation factors into $(D + 8)(D - 4)y = 0.$
So our roots are -8 and 4.

*Step 3: Write the general solution.*

$$ y(x) = c_1\exp(-8x) + c_2\exp(4x) $$ 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) = -8c_1\exp(-8x) + 4c_2\exp(4x).$ Then we plug in $ x=0 $ to get the following equations. $$ \begin{align} y(0) = c_1 + c_2 &= -8 \\ y'(0) = -8c_1 + 4c_2 &= 6 \end{align} $$ We solve these equations to get $ c_1= -19/6 $ and $ c_2 = -29/6.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.

$$
y(x) = -(19/6)\exp(-8x) - (29/6)\exp(4x)
$$
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©2010, 2014 Andrew G. Bennett