Find the general solution of $ y'''- y'' - 9y' +9y = 0 $ Find the general solution of  $ y'''- y'' - 9y' +9y = 0 $
The answer is
$y=c_{1}e^{-3x}+c_{2}e^{3x}+c_{3}e^{x}$
how do i approach this problem?
 A: There is a "universal" approach to this type of problem. We look for solutions of the form $e^{mx}$. Substituting $y=e^{mx}$ into our DE, we get after a short while $m^3e^{mx}-m^2e^{mx}-9me^{mx}+9e^{mx}=0$.
This will be satisfied if and only if $m^3-m^2-9m+9=0$.
The above cubic equation happens to be easy to solve. It can be rewritten as $m^2(m-1)-9(m-1)=0$, and then as $(m-1)(m^2-9)=0$. The solutions are $m=1$, $m=3$, and $m=-3$.
So $e^x$, $e^{3x}$, and $e^{-3x}$ are solutions of our DE. Because our DE is linear and right-hand side is $0$, that means that for any constants $A$, $B$, and $C$, $y=Ae^x+Be^{3x}+Ce^{-3x}$ is a solution.
Because our differential equation has order $3$, and (it turns out) the $3$ solutions we found are linearly independent, all solutions are of the form $y=Ae^x+Be^{3x}+Ce^{-3x}$. 
Remark: The above is a more or less universal method for dealing with linear differential equations that have constant coefficients and right-hand side $0$. There are some complications. If the polynomial equation that we get has multiple roots (example: $(m-1)(m-3)^2$) then some adjustment needs to be made. It turns out that in that example, $e^x$, $e^{3x}$, and $xe^{3x}$ will give us a set of basic solutions.
Also, if some of the roots of the polynomial we get are non-real, we proably want to make an adjustment, and express a solution like $y=e^{5ix}$, which is technically correct, in terms of sines and cosines. 
A: Hint: use characteristic equation $$y'''- y'' - 9y' +9y = 0 $$ $$(D^3-D^2-9D+9)y=0$$ then $$(D-1)(D-3)(D+3)y=0$$we have $D=1,3,-3$ and $$y=c_{1}e^{-3x}+c_{2}e^{3x}+c_{3}e^{x}$$
A: Try to substitute $e^{\lambda x}$ in the equation and solve the algebraic equation in $\lambda$ as is usually done for second order homogeneous ODEs.
A: You have the following differential equation:
$$ y^{'''}-y^{''}-9y^{'}+9y=0 $$
Applying the Laplace transform:
$$ Y(s)(s^{3}-s^{2}-9s+9)=K_{1}+sK_{2}+s^{2}K_{3} $$
Rearranging:
$$ Y(s)=\frac{K_{1}+sK_{2}+s^{2}K_{3}}{(s+3)(s-3)(s-1)} $$
I can decompose the division into simple fractions:
$$ Y(s)=\frac{C_{1}}{s+3}+\frac{C_{2}}{s-3}+\frac{C_{1}}{s-1}$$
Doing the inverse laplace transform:
$y=C_{1}e^{-3x}+C_{2}e^{3x}+C_{3}e^{x}$
