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. $$ (D^2 + 9)y = 0 $$ Substep 2: Find the roots. We use the quadratic formula to compute the roots, $ \pm 3i$

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

Since we already have a root of 0, 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 $ 9Ax + 9B = 5x + 5.$ This gives us the equations $9A=5$ and $9B=5.$ From these equations we get $A=5/9$ and $B=5/9.$ So we get $$ y_p(x)=(5/9)x + 5/9. $$ 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 $$ y(x) = (5/9)x + 5/9 + c_1\cos(3x) + c_2\sin(3x) $$ 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) = 5/9 - 3c_1sin(3x) + 3c_2\cos(3x).$ Then we plug in $ x=0$ to get the following equations. $$ \begin{align} y(0) = 5/9 + c_1 &= -6 \\y'(0) = 5/9 + 3c_2 &= 9 \end{align} $$ We solve these equations to get $ c_1=-59/9$ and $ c_2=76/27.$ Finally, we plug our values for $ c_1$ and $ c_2$ into the general solution to find our solution to the initial value problem. $$ y(x) = (5/9)x + 5/9 - (59/9)\cos(3x) + (76/27)\sin(3x). $$ You may reload this page to generate additional examples.