Case #2 involves the situation where our characteristic equation results in 1 real repeated root and will involve the use of D'Alembert's technique, which was explained in this example.
(Ex 1) Solve the initial value problem for the given 2nd order homogeneous differential equation
$$y''-2y'+y=0 $$
...where:
$$y(0)=5 \;,\; y'(0)=10 $$
Using our previously described substitutions, let:
$$y=e^{rx} $$
$$y' = re^{rx} $$
$$y'' = r^2e^{rx} $$
... and plug into our given differential equation:
$$r^2e^{rx}-2re^{rx}+e^{rx}=0 $$
Now proceed to obtain our "characteristic equation":
$$e^{rx}(r^2-2r+1)=0 $$
$$(r-1)(r-1)=0 $$
As you can see, we have one real repeated root:
$$r_1=1 $$
...which means that one solution to our 2nd order differential equation when back-substituting for y is:
$$y_1 = e^{r_1x} $$
$$y_1 = e^x $$
Now recall that given a single solution to a second order differential equation, we can use D'Alembert's procedure to find a second solution y2 where:
$$y_2 = vy_1 $$
Doing so involves making a substitution (v), where:
$$v' = e^{-\int{}{}(\frac{2y'_1}{y_1}+p)dx} $$
In our example:
$$y_1 = e^x $$
$$y'_1 = e^x $$
$$p = -2 $$
...which gives us:
$$v' = e^{-\int{}{}(2(1)-2)dx}$$
$$v' = e^0 $$
$$v' = 1 $$
$$v = \int{}{}1 \; dx $$
$$v=x $$
Therefore:
$$y_2 = vy_1 $$
$$y_2 = xe^x$$
This gives us a fundamental set of:
$$e^x \;,\; xe^x $$
Also, recall that any solution multiplied by a constant is also a solution, as well as the fact that when given two solutions to a second order differential equation, the sum of those solutions is also a solution. This gives the following general solution to the differential equation:
$$y_g = C_1e^x + C_2xe^x $$
Now use the given initial conditions to find the particular solution to our initial value problem:
$$y(0) = C_1e^0+C_2(0)e^0=5 $$
$$C_1=5 $$
and
$$y'(x) = \frac{d}{dx}(C_1e^x + C_2xe^x) $$
$$y'(x) = C_1e^x + C_2xe^x + C_2e^x$$
$$y'(x) = 5e^x + C_2(xe^x+e^x) $$
$$y'(0) = 5+C_2(0+1) = 10 $$
$$5+C_2=10 $$
$$C_2=5 $$
Now we substitute our above-determined values for C1 and C2 into our general solution:
$$y_p = 5e^x+5xe^x $$
We have now found the particular solution (yp) to our initial value problem
Continue on to case #3 (complex roots of characteristic equation)