(Ex 1) Solve the initial value problem for the given 2nd order homogeneous differential equation:
$$y''-4y'+3y=0 $$
where:
$$y(0)=7 \;,\; y'(0)=11 $$
Now use the following substitutions:
$$y=e^{rx} $$
$$y'=re^{rx} $$
$$y'' = r^2e^{rx} $$
...and plug them into our equation:
$$r^2e^{rx}-4re^{rx}+3e^{rx}=0 $$
Proceed to obtain our "characteristic equation":
$$e^{rx}(r^2-4r+3)=0 $$
$$r^2-4r+3=0 $$
...and solve for r:
$$(r-3)(r-1)=0 $$
$$r_1=3 \;,\; r_2=1 $$
Back-substituting these values for "r" into our equation for "y" (and recalling that any solution multiplied by a constant is also a solution) gives us :
$$y_1=C_1e^{3x} \;,\; y_2=C_2e^x $$
Furthermore, if two functions are solutions to a second order differential equation, then their sum is also a solution. Therefore, general solutions to our 2nd order differential equation are represented by:
$$y_g = C_1e^{3x} + C_2e^x $$
Now proceed to solve for the initial value problem:
$$y(0)=C_1e^0+C_2e^0=7 $$
$$C_1+C_2=7 \qquad(1) $$
and
$$y'(0)=3C_1e^0+C_2e^0=11 $$
$$3C_1+C_2=11 \qquad(2) $$
From equation #1:
$$C_1=7-C_2 \qquad(3)$$
Substituting eqn 3 into eqn 2:
$$21-3C_2+C_2=11 $$
$$21-2C_2=11 $$
$$C_2=5 $$
Substituting 5 for C2 in equation 1:
$$C_1 + 5 =7 $$
$$C_1 = 2 $$
Now that we have solved for our constants, we can plug them into our general solution and obtain a particular solution for the given initial value problem:
$$y_p = 2e^{3x}+5e^x $$
Continue on to case #2 (r1 = r2, 1 real repeated root)