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### Second-Order Linear Inhomogeneous Equations

#### Additional Examples

Solve the initial value problem
$$
\begin{align}
\frac{d^2y}{dx^2} + 3\frac{dy}{dx} - 18y &= 6\exp(-7x) \\
y(0) &= 8 \\
y'(0) &= 9
\end{align}
$$

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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.*
$$
(D^2 + 3D - 18)y = 0
$$
*Substep 2: Find the roots.*
The equation factors into $ (D + 6)(D - 3)y = 0.$
So our roots are -6 and 3.

*Substep 3: Write the homogeneous solutions.*
$$
y_h(x) = c_1\exp(-6x) + c_2\exp(3x)
$$
__Step 2: Guess the form of the particular solutions__

Since the right-hand side is $6\exp(-7x),$ 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 $ 10A\exp(-7x) = 6\exp(-7x).$ This gives us the equation $10A=6,$ from which we get $ A=3/5.$ So we get
$$
y_p(x)=(3/5)\exp(-7x).
$$
__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
$$
y(x) = (3/5)\exp(-7x) + c_1\exp(-6x) + c_2\exp(3x)
$$
**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) = -(21/5)\exp(-7x) - 6c_1\exp(-6x) + 3c_2\exp(3x).$
Then we plug in $ x=0$ to get the following equations.
$$
\begin{align}
y(0) = 3/5 + c_1 + c_2 &= 8 \\y'(0) = -21/5 - 6c_1 + 3c_2 &= 9
\end{align}
$$
We solve these equations to get $ c_1=1$ and $ c_2=32/5.$
Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem.
$$
y(x) = (3/5)\exp(-7x) + \exp(-6x) + (32/5)\exp(3x).
$$
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©2010, 2014 Andrew G. Bennett