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### Second-Order Linear Inhomogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 29y &= -2\exp(7x) \\
y(0) &= -4 \\
y'(0) &= 1
\end{align}
$$

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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: Find the homogeneous solution.__

*Substep 1: Write the equation in operator form.*
$$
(D^2 + 4D + 29)y = 0
$$
*Substep 2: Find the roots.*
We use the quadratic formula to compute the roots,
$
-2\pm 5i$

*Substep 3: Write the homogeneous solutions.*
$$
y_h(x) = c_1\exp(-2x)\cos(5x) + c_2\exp(-2x)\sin(5x)
$$
__Step 2: Guess the form of the particular solutions__

Since the right-hand side is $-2\exp(7x),$ we guess $ y_p(x) = A\exp(7x).$

__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 $ 106A\exp(7x) = -2\exp(7x).$ This gives us the equation $106A=-2,$ from which we get $ A=-1/53.$ So we get
$$
y_p(x)=-(1/53)\exp(7x).
$$
__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
$$
y(x) = -(1/53)\exp(7x) + c_1\exp(-2x)\cos(5x) + c_2\exp(-2x)\sin(5x)
$$
**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) = -(7/53)\exp(7x) - 2c_1\exp(-2x)\cos(5x) - 5c_1\exp(-2x)\sin(5x) - 2c_2\exp(-2x)\sin(5x) + 5c_2\exp(-2x)\cos(5x).$
Then we plug in $ x=0$ to get the following equations.
$$
\begin{align}
y(0) = -1/53 + c_1 &= -4 \\y'(0) = -7/53 - 2c_1 + 5c_2 &= 1
\end{align}
$$
We solve these equations to get $ c_1=-211/53$ and $ c_2=-362/265.$
Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.
$$
y(x) = -(1/53)\exp(7x) - (211/53)\exp(-2x)\cos(5x) - (362/265)\exp(-2x)\sin(5x).
$$
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©2010, 2014 Andrew G. Bennett