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### Second-Order Linear Homogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 6y &= 0 \\
y(0) &= -2 \\
y'(0) &= 10
\end{align}
$$

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©2010 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.*
$$
(D^2 + 5D - 6)y = 0
$$
*Step 2: Find the roots.*
The equation factors into $(D + 6)(D - 1)y = 0.$
So our roots are -6 and 1.

*Step 3: Write the general solution.*

$$ y(x) = c_1\exp(-6x) + c_2\exp(x) $$ 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) = -6c_1\exp(-6x) + c_2\exp(x).$ Then we plug in $ x=0 $ to get the following equations. $$ \begin{align} y(0) = c_1 + c_2 &= -2 \\ y'(0) = -6c_1 + c_2 &= 10 \end{align} $$ We solve these equations to get $ c_1= -12/7 $ and $ c_2 = -2/7.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.

$$
y(x) = -(12/7)\exp(-6x) - (2/7)\exp(x)
$$
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©2010 Andrew G. Bennett