Kansas State University

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. $$ (D^2 + 8D + 25)y = 0 $$ Substep 2: Find the roots. We use the quadratic formula to compute the roots, $ -4\pm 3i$

Substep 3: Write the homogeneous solutions. $$ y_h(x) = c_1\exp(-4x)\cos(3x) + c_2\exp(-4x)\sin(3x) $$ Step 2: Guess the form of the particular solutions

Since the right-hand side is $3\exp(-4x),$ we guess $ y_p(x) = A\exp(-4x).$

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 $ 9A\exp(-4x) = 3\exp(-4x).$ This gives us the equation $9A=3,$ from which we get $ A=1/3.$ So we get $$ y_p(x)=(1/3)\exp(-4x). $$ 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 $$ y(x) = (1/3)\exp(-4x) + c_1\exp(-4x)\cos(3x) + c_2\exp(-4x)\sin(3x) $$ 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) = -(4/3)\exp(-4x) - 4c_1\exp(-4x)\cos(3x) - 3c_1\exp(-4x)\sin(3x) - 4c_2\exp(-4x)\sin(3x) + 3c_2\exp(-4x)\cos(3x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = 1/3 + c_1 &= -5 \\y'(0) = -4/3 - 4c_1 + 3c_2 &= 3 \end{align} $$ We solve these equations to get $ c_1=-16/3$ and $ c_2=-17/3.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = (1/3)\exp(-4x) - (16/3)\exp(-4x)\cos(3x) - (17/3)\exp(-4x)\sin(3x). $$ You may reload this page to generate additional examples.