Math 240 Home, Textbook Contents, Online Homework Home

Warning: MathJax requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

Second-Order Linear Inhomogeneous Equations

Additional Examples

Solve the initial value problem $$ \begin{align} \frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 13y &= 6\exp(-6x) \\ y(0) &= -2 \\ y'(0) &= -2 \end{align} $$

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 $6\exp(-6x),$ we guess $ y_p(x) = A\exp(-6x).$

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 $ 25A\exp(-6x) = 6\exp(-6x).$ This gives us the equation $25A=6,$ from which we get $ A=6/25.$ So we get $$ y_p(x)=(6/25)\exp(-6x). $$ 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 $$ y(x) = (6/25)\exp(-6x) + c_1\exp(-2x)\cos(3x) + c_2\exp(-2x)\sin(3x) $$ 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) = -(36/25)\exp(-6x) - 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. $$ \begin{align} y(0) = 6/25 + c_1 &= -2 \\y'(0) = -36/25 - 2c_1 + 3c_2 &= -2 \end{align} $$ We solve these equations to get $ c_1=-56/25$ and $ c_2=-42/25.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = (6/25)\exp(-6x) - (56/25)\exp(-2x)\cos(3x) - (42/25)\exp(-2x)\sin(3x). $$ You may reload this page to generate additional examples.


If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010, 2014 Andrew G. Bennett