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Second-Order Linear Inhomogeneous Equations

Additional Examples

Solve the initial value problem $$ \begin{align} \frac{d^2y}{dx^2} + 8\frac{dy}{dx} - 20y &= -3\exp(-7x) \\ y(0) &= -9 \\ y'(0) &= -1 \end{align} $$

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. $$ (D^2 + 8D - 20)y = 0 $$ Substep 2: Find the roots. The equation factors into $ (D + 10)(D - 2)y = 0.$ So our roots are -10 and 2.

Substep 3: Write the homogeneous solutions. $$ y_h(x) = c_1\exp(-10x) + c_2\exp(2x) $$ Step 2: Guess the form of the particular solutions

Since the right-hand side is $-3\exp(-7x),$ 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 $ -27A\exp(-7x) = -3\exp(-7x).$ This gives us the equation $-27A=-3,$ from which we get $ A=1/9.$ So we get $$ y_p(x)=(1/9)\exp(-7x). $$ 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 $$ y(x) = (1/9)\exp(-7x) + c_1\exp(-10x) + c_2\exp(2x) $$ 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) = -(7/9)\exp(-7x) - 10c_1\exp(-10x) + 2c_2\exp(2x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = 1/9 + c_1 + c_2 &= -9 \\y'(0) = -7/9 - 10c_1 + 2c_2 &= -1 \end{align} $$ We solve these equations to get $ c_1=-3/2$ and $ c_2=-137/18.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = (1/9)\exp(-7x) - (3/2)\exp(-10x) - (137/18)\exp(2x). $$ You may reload this page to generate additional examples.


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©2010, 2014 Andrew G. Bennett