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### Second-Order Linear Inhomogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 13y &= -3\exp(x) \\
y(0) &= 3 \\
y'(0) &= 2
\end{align}
$$

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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: Find the homogeneous solution.__

*Substep 1: Write the equation in operator form.*
$$
(D^2 + 4D + 13)y = 0
$$
*Substep 2: Find the roots.*
We use the quadratic formula to compute the roots,
$
-2\pm 3i$

*Substep 3: Write the homogeneous solutions.*
$$
y_h(x) = c_1\exp(-2x)\cos(3x) + c_2\exp(-2x)\sin(3x)
$$
__Step 2: Guess the form of the particular solutions__

Since the right-hand side is $-3\exp(x),$ we guess $ y_p(x) = A\exp(x).$

__Step 3: Plug our guess into the equation and solve for the undetermined coefficients__

Plugging our guess for $ y_p(x)$ into the equation, we obtain $ 18A\exp(x) = -3\exp(x).$ This gives us the equation $18A=-3,$ from which we get $ A=-1/6.$ So we get
$$
y_p(x)=-(1/6)\exp(x).
$$
__Step 4: The general solution is the particular solution plus all the homogeneous solutions.__

So from the results of steps 1 and 3, we get the general solution is
$$
y(x) = -(1/6)\exp(x) + c_1\exp(-2x)\cos(3x) + c_2\exp(-2x)\sin(3x)
$$
**Second, we solve for the constants**

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) = -(1/6)\exp(x) - 2c_1\exp(-2x)\cos(3x) - 3c_1\exp(-2x)\sin(3x) - 2c_2\exp(-2x)\sin(3x) + 3c_2\exp(-2x)\cos(3x).$
Then we plug in $ x=0$ to get the following equations.
$$
\begin{align}
y(0) = -1/6 + c_1 &= 3 \\y'(0) = -1/6 - 2c_1 + 3c_2 &= 2
\end{align}
$$
We solve these equations to get $ c_1=19/6$ and $ c_2=17/6.$
Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.
$$
y(x) = -(1/6)\exp(x) + (19/6)\exp(-2x)\cos(3x) + (17/6)\exp(-2x)\sin(3x).
$$
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©2010 Andrew G. Bennett