Is there any alternative to Wronskian? Calculating a Wronskian is a very painful process, especially for higher order differential equations. Actually, I'm trying to solve a 4th order non-homogeneous linear differential equation. Consider the equation given below, for example,
$$    y'''' - u^4y = e^{ux}, \quad \text{where $u > 0$ is a constant.} $$
I got general solution,
$$    y = C_1.e^{ux} + C_2.e^{-ux} + C_3.cos(ux) + C_4.sin(ux)$$
to homogeneous equation, 
$$    y'''' - u^4y = 0, \quad \text{where $u > 0$ is a constant.} $$
As we can see, we got four functions, namely, $f1...f4$, to calculate Wronskian. Here I ended up solving a $4X4$ determinant.
Is there any alternative?
Note: I solved it using undeteremined coefficient, $y_p = x.p.e^{ux}$. Still your answers were helpful. Thanks.
 A: It is easy to find a particular solution to $y^{(4)}-u^4y=e^{ux}$, so you can focus on the linear equation $y^{(4)}-u^4y=0$.
If $X=(y''',y'',y',y)$, then $X'=AX$ with $A=\left( \begin{array}{cccc} 0&0&0&u^4 \\ 1 & 0&0&0 \\ 0&1&0&0 \\ 0&0&1&0  \end{array} \right)$. You only have to compute $\exp(A)$ (notice that $A^4= u^4 \operatorname{Id}$).
A: There are many ways to approach it.
For example, for the homogeneous case, we could have written the characteristic equation as:
$$m^4 -u^4 = 0$$
This leads to the four roots $\pm u$ and $\pm iu$.
From that knowledge, we can write the solution as:
$$y_H = y = C_1.e^{ux} + C_2.e^{-ux} + C_3.\cos ux + C_4.\sin ux$$
Now, we need only deal with the particular solution, and we see that one of the roots is coincident with the particular (RHS) function, so there are pretty straightforward ways to deal with that.
Another approach is to linearize the system and write it as $X' = AX$ as was shown in another answer.
Here, you will find the same roots to the characteristic polynomial, can diagnoalize the system (if it is diagonalizable) and write the exponential solution for that matrix.
There are other ways too, but these two are worth exploring.
A: Given an $n$-th order linear homogeneous differential equation
$$y^{(n)} +p_{n-1}(x)y^{(n-1)} +\ldots + p_1(x) y'+ p_0(x) y=0\tag{1}$$
the Wronskian is a quantity $t\mapsto W(t):=W[y_1,\ldots, y_n](t)$ associated to a basis ${\bf b}=(y_1,\ldots, y_n)$ of the solution space ${\cal L}$. Formally it is the determinant of the matrix formed by the jet-extensions $(y_k, y_k', \ldots, y_k^{(n-1)})$ of the basis functions. From this one can conclude that $W(\cdot)$ is multiplied with a constant factor when the basis ${\bf b}$ is replaced by some other basis ${\bf b}'$, and this in turn suggests that $W(\cdot)$ satisfies a certain linear differential equation of the form $w'= c(x) w$, where $c(x)$ can be determined directly from $(1)$. In fact Abel's identity says that
$$W'(x)=-p_{n-1}(x) W(x)\ .$$
In the case at hand it follows that $W(\cdot)$ is a constant. The numerical value of this constant depends on the chosen basis ${\bf b}$, and there is no simple way out of computing it, other than putting $x=0$.
