Second Order homogeneous differential equation with two unique real roots

(Ex 1) Solve the initial value problem for the given 2nd order homogeneous differential equation:

$y^{″} - 4 y^{'} + 3 y = 0$

where:

$y (0) = 7, y^{'} (0) = 11$

Now use the following substitutions:

$y = e^{r x}$

$y^{'} = r e^{r x}$

$y^{″} = r^{2} e^{r x}$

...and plug them into our equation:

$r^{2} e^{r x} - 4 r e^{r x} + 3 e^{r x} = 0$

Proceed to obtain our "characteristic equation":

$e^{r x} (r^{2} - 4 r + 3) = 0$

$r^{2} - 4 r + 3 = 0$

...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 :

$y_{1} = C_{1} e^{3 x}, y_{2} = C_{2} e^{x}$

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:

$y_{g} = C_{1} e^{3 x} + C_{2} e^{x}$

Now proceed to solve for the initial value problem:

$y (0) = C_{1} e^{0} + C_{2} e^{0} = 7$

$C_{1} + C_{2} = 7 (1)$

and

$y^{'} (0) = 3 C_{1} e^{0} + C_{2} e^{0} = 11$

$3 C_{1} + C_{2} = 11 (2)$

From equation #1:

$C_{1} = 7 - C_{2} (3)$

Substituting eqn 3 into eqn 2:

$21 - 3 C_{2} + C_{2} = 11$

$21 - 2 C_{2} = 11$

$C_{2} = 5$

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:

$y_{p} = 2 e^{3 x} + 5 e^{x}$

Continue on to case #2 (r1 = r2, 1 real repeated root)