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### Second-Order Linear Inhomogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} - 7\frac{dy}{dx} &= -2x \\
y(0) &= 3 \\
y'(0) &= -4
\end{align}
$$

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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.*
$$
(D^2 - 7D)y = 0
$$
*Substep 2: Find the roots.*
The equation factors into $ D(D - 7)y = 0.$
So our roots are 0 and 7.

*Substep 3: Write the homogeneous solutions.*
$$
y_h(x) = c_1 + c_2\exp(7x)
$$
__Step 2: Guess the form of the particular solutions__

Since the right-hand side is $-2x,$ we guess $ y_p(x) = Ax^2 + Bx.$

__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 $ -14Ax - 7B + 2A = -2x.$ This gives us the equations $-14A=-2$ and $ - 7B + 2A=0.$ From these equations we get $A=1/7$ and $B=2/49.$ So we get
$$
y_p(x)=(1/7)x^2 + (2/49)x.
$$
__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
$$
y(x) = (1/7)x^2 + (2/49)x + c_1 + c_2\exp(7x)
$$
**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) = (2/7)x + 2/49 + 7c_2\exp(7x).$
Then we plug in $ x=0$ to get the following equations.
$$
\begin{align}
y(0) = c_1 + c_2 &= 3 \\y'(0) = 2/49 + 7c_2 &= -4
\end{align}
$$
We solve these equations to get $ c_1=1227/343$ and $ c_2=-198/343.$
Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.
$$
y(x) = (1/7)x^2 + (2/49)x + 1227/343 - (198/343)\exp(7x).
$$
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©2010, 2014 Andrew G. Bennett