We can replace this with

r

r

(r - 3) ( r

r = 3, +- (6)

The complex solutions result in a different way, resulting in sin and cosine solutions. This can be proven using complex analysis and substituting in using euler's equation.^{3}- 3 r^{2}+ 6 r - 18 = 0
We then solve this equation in order to find the coefficients to a general solution.

^{2}(r - 3) + 6( r - 3) = 0(r - 3) ( r

^{2}+ 6) = 0r = 3, +- (6)

^{1/2}i
So now that we solved this equation, we can note that theres a 3. This means that the coefficient of an exponential is 3

y = c

_{1}e^{3x}^{}
Resulting in a final combined solution of this

Which is in a form we can use in order to completely solve if we have initial conditions to find the coefficients c

_{1}, c_{2}and c_{3}. Lets use the example that at y(0) = 1, y'(0) = 1, y''(0) = 1. This would mean that
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