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

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

Since the right-hand side is $-5x + 4,$ we guess $ y_p(x) = Ax + B.$

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 $ 8Ax + 8B + 4A = -5x + 4.$ This gives us the equations $8A=-5$ and $8B + 4A=4.$ From these equations we get $A=-5/8$ and $B=13/16.$ So we get $$ y_p(x)=-(5/8)x + 13/16. $$ 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 $$ y(x) = -(5/8)x + 13/16 + c_1\exp(-2x)\cos(2x) + c_2\exp(-2x)\sin(2x) $$ 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) = -5/8 - 2c_1\exp(-2x)\cos(2x) - 2c_1\exp(-2x)\sin(2x) - 2c_2\exp(-2x)\sin(2x) + 2c_2\exp(-2x)\cos(2x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = 13/16 + c_1 &= 9 \\y'(0) = -5/8 - 2c_1 + 2c_2 &= 10 \end{align} $$ We solve these equations to get $ c_1=131/16$ and $ c_2=27/2.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = -(5/8)x + 13/16 + (131/16)\exp(-2x)\cos(2x) + (27/2)\exp(-2x)\sin(2x). $$ You may reload this page to generate additional examples.