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### Second-Order Linear Inhomogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} - \frac{dy}{dx} &= 6\sin(x) \\
y(0) &= 9 \\
y'(0) &= 4
\end{align}
$$

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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.*
$$
(D^2 - D)y = 0
$$
*Substep 2: Find the roots.*
The equation factors into $ D(D - 1)y = 0.$
So our roots are 0 and 1.

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

Since the right-hand side is $6\sin(x),$ we guess $ y_p(x) = A\cos(x) + B\sin(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 $ (-A - B)\cos(x) + (A - B)\sin(x) = 6\sin(x).$ This gives us the equations $-A - B=0$ and $A - B=6.$ From these equations we get $A=3$ and $B=-3.$ So we get
$$
y_p(x)=3\cos(x) - 3\sin(x).
$$
__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
$$
y(x) = 3\cos(x) - 3\sin(x) + c_1 + c_2\exp(x)
$$
**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) = -3\cos(x) - 3\sin(x) + c_2\exp(x).$
Then we plug in $ x=0$ to get the following equations.
$$
\begin{align}
y(0) = 3 + c_1 + c_2 &= 9 \\y'(0) = -3 + c_2 &= 4
\end{align}
$$
We solve these equations to get $ c_1=-1$ and $ c_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.
$$
y(x) = 3\cos(x) - 3\sin(x) - 1 + 7\exp(x).
$$
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©2010 Andrew G. Bennett