How to solve 3rd order Ordinary Differential Equation by using Wronskian? The ODE is  $$ (D^3-6D^2+11D-6)y=e^{2x}$$
I know how to solve $2$nd order. But, I don't know the formula for $3$rd order.
let the complementary function be
$$y=c_1f_1+c_2f_2+c_3f_3$$
let the particular integral be 
$$y=c_1(x)f_1+c_2(x)f_2+c_3(x)f_3$$
After solving the $3$ degree matrix I get the value of $W(f_1,f_2,f_3)$
But what to do next ? What is the formula to find   the value of $c_1(x),c_2(x),c_3(x)$
 A: I guess you mean method of variation of parameters for a third Order Differential Equation. And you already have solved the homogeneous equation.
It's $$\pmatrix { c'_1 \\ c'_2 \\c'_3}=W^{-1}\pmatrix { 0 \\ 0 \\e^{2x} }$$
Where $W^{-1}$ is the inverse of Wronskian matrix  and $c_i'$ are derivatives.

Edit I added some calculations
$$W=\pmatrix {e^x & e^{2x} &e^{3x} \\ e^x & 2e^{2x} & 3e^{3x} \\ e^x & 4e^{2x} & 9e^{3x}}$$
it's easy to find the determinant $|W|=2e^{6x}$
Now you need the inverse of the Wronskian Matrix ...We don't need all the elements of the matrix. Only the last column is needed:
$$W^{-1}=\frac 1 {|W|}\pmatrix {* & * &e^{5x} \\ * & *& -2e^{4x} \\ * & * & e^{3x}}$$
So we have that:
$$\pmatrix { c'_1 \\ c'_2 \\c'_3}=\frac 1 {|W|}\pmatrix {* & * &e^{5x} \\ * & *& -2e^{4x} \\ * & * & e^{3x}}\pmatrix { 0 \\ 0 \\e^{2x} }$$
$$\pmatrix { c'_1 \\ c'_2 \\c'_3}=\frac 1 {|W|}\pmatrix { e^{7x}\\ -2e^{6x} \\e^{5x} }$$
$$\pmatrix { c'_1 \\ c'_2 \\c'_3}=\frac 1 {2}\pmatrix { e^{x}\\ -2 \\e^{-x} }$$
Now integrate to find the $c_i$ coefficients:
$$\pmatrix { c_1 \\ c_2 \\c_3}=\pmatrix { \frac 1 {2}e^{x}\\ -x \\-\frac 1 {2}e^{-x} }$$
The particular solution is therefore:
$$y_p=c_1e^x+c_2e^{2x}+c_3e^{3x}$$
$$y_p=\frac 1 {2}e^{2x}-xe^{2x}-\frac 1 {2}e^{2x}$$
$$y_p=-xe^{2x}$$
Variation of Constants is a much better method for your differential equation. Your guess should be: $$y_p=Axe^{2x}$$
A: $$(D^3-6D^2+11D-6)y=e^{2x}\tag1$$
Auxiliary equation is $~m^3-6m^2+11m-6=0\implies (m-1)(m-2)(m-3)=0$.
Roots of the auxiliary equation is $~m=1,~2,~3~.$
So complementary function (C.F.)$~~~=c_1 ~e^x~+~c_2~e^{2x}~+~c_3~e^{3x}~$, where $~c_1,~c_2,~c_3~$are constants.
Here $~f(D)=D^3-6D^2+11D-6~$.
Clearly, $~f(2)=0~$but$~f'(2)=-1\ne 0~$.
So particular integral (P.I.),
P.I.$~~=\dfrac{1}{D^3-6D^2+11D-6}~e^{2x}$
$~~~~~~~=x~\dfrac{e^{2x}}{-1}~~~~~($ by rule $2~)$
$~~~~~~~=-~x~e^{2x}$
So the general solution of the differential equation $(1)$ is
$$y(x)=~\text{C.F.}~+~\text{P.I.}$$
$$\implies y(x)=c_1 ~e^x~+~c_2~e^{2x}~+~c_3~e^{3x}~-~x~e^{2x}$$
where $~c_1,~c_2,~c_3~$are constants.


Consider a differential equation of the form $f(D)y=X$
If $X=e^{ax}$, then
$1.$ P.I.$\quad = \frac{1}{f(D)}e^{ax}=\frac{e^{ax}}{f(a)}$, if $f(a)\neq 0$
$2.$ P.I.$\quad =\frac{1}{f(D)}e^{ax}=x~\frac{e^{ax}}{f'(a)}$, if $~f(a)=0~$but$~f'(a)\ne 0$
$3.$ P.I.$\quad =\frac{1}{f(D)}e^{ax}=x^2~\frac{e^{ax}}{f''(a)}$, if $~f(a)=0,~f'(a)= 0~$but$~f''(a)\ne 0$
and so on.

A: The formula for variation of parameter is the same regardless of order. This is because they all come from writing the equation as a first order vector-valued ODE, by making vector of the form $(f(x),f^{\prime}(x),...,f^{(n)}(x))^{t}$
For any first order vector-valued homogeneous ODE: $v^{\prime}(x)=A(x)v(x)$ where $A$ is a square $n\times n$ matrix, then the Wronskian matrix is a square matrix solution that is non-singular at one point: some square $W(x)$ such that $W^{\prime}(x)=A(x)v(x)$.
Assuming you solved $W$, the inhomogeneous version can be solved by variation of parameter. We have the inhomogeneous equation $v^{\prime}(x)-A(x)v(x)=b(x)$. Writing $D$ for $\frac{d}{dx}$ and writing the equation without the argument $(x)$ then this become $Dv-Av=b$. Since $Dv-Av=D(WW^{-1}v)-A(WW^{-1}v)=(DW)(W^{-1}v))+WD(W^{-1}v)-(AW)(W^{-1}v)=(DW-AW)(W^{-1}v)+WD(W^{-1}v)$. Making use of the definition of $W$ then $DW-AW=0$ so $WD(W^{-1}v)=b$ so $D(W^{-1}v)=W^{-1}b$. This is the formula for variation of parameter.
