Find the general solution of $y''' - 5y'' -22y' + 56y = 0$, $y(0) = 1$, $y'(0) = -2$, $y"(0) = -4$
I need help finding the initial condition and I have no clue how to find it. I get stuck after I substitute "t" and "y" on the general solution.
Solution:
$y''' - 5y'' - 22y' + 56y = 0 \tag{1}$
$$r^3 - 5r^2 - 22r + 56 = 0$$
$$y(t) = C_1(e^{-4t}) + C_2(e^{2t}) + C_3(e^{7t}) $$
Now this is where I get stuck finding the initial condition.
$$1 = y(0) = C_1 + C_2 + C_3$$
$$-2 = y'(0) = -4C_1 + 2C_2 + 7C_3$$
$$-4 = y"(0) = 16C_1 + 4C_2 + 49C_3$$
Can someone help me on how to find the initial condition. Thanks!
You have a system of three linear equations in three unknowns. There are several ways to solve this. I would write it as an augmented matrix and do row-reduction, but you can also solve the first equation for $C_3$ and plug that into the second and third equations. Then solve the second equation for $C_2$ and plug it into the third equation. Then solve the third equation for $C_1.$ Then use your previous work to find $C_2$ and $C_3$.