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. $$ (D^2 + 4D + 20)y = 0 $$ Substep 2: Find the roots. We use the quadratic formula to compute the roots, $ -2\pm 4i$

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

Since the right-hand side is $-\cos(2x),$ we guess $ y_p(x) = A\cos(2x) + B\sin(2x).$

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 $ (16A + 8B)\cos(2x) + (-8A + 16B)\sin(2x) = -\cos(2x).$ This gives us the equations $16A + 8B=-1$ and $-8A + 16B=0.$ From these equations we get $ A=-1/20$ and $B=-1/40.$ So we get $$ y_p(x)=-(1/20)\cos(2x) - (1/40)\sin(2x). $$ 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 $$ y(x) = -(1/20)\cos(2x) - (1/40)\sin(2x) + c_1\exp(-2x)\cos(4x) + c_2\exp(-2x)\sin(4x) $$ 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) = -(1/20)\cos(2x) + (1/10)\sin(2x) - 2c_1\exp(-2x)\cos(4x) - 4c_1\exp(-2x)\sin(4x) - 2c_2\exp(-2x)\sin(4x) + 4c_2\exp(-2x)\cos(4x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = -1/20 + c_1 &= -6 \\y'(0) = -1/20 - 2c_1 + 4c_2 &= -8 \end{align} $$ We solve these equations to get $ c_1=-119/20$ and $ c_2=-397/80.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = -(1/20)\cos(2x) - (1/40)\sin(2x) - (119/20)\exp(-2x)\cos(4x) - (397/80)\exp(-2x)\sin(4x). $$ You may reload this page to generate additional examples.