# Particular solution to 4th order ode

1. May 3, 2012

### physicsjock

hey,

i have this 4th order ode question that i've been working on,

the homogeneous solution was easy enough by finding the particular solution has become a bit annoying,

the ode is

y'''' - 4y'' = 5x2 - e2x

I have gotten the particular solution using variation of parameters, but I used mathematica to do the work, like finding the inverse of the wronskian matrix and integrating the ugly results.

the particular solution i get with mathematica is

yp=1/192 (-30 - 3 E^(2 x) (-5 + 4 x) - 20 x^2 (3 + x^2))

I have checked that this is the particular solution by substitution into the ode and it gives required result.

My question is,

there is no way I was supposed to do this using mathematica, is there a way to approximate the particular solution, like with 2nd order odes?

I tried that in this case,

my particular solution I tried was

(A + Bx + Cx2)+De2x

which is just the sum of the particular solution of x^2 and e^(2x)

Is there some way to estimate the particular solution without using variation of parameters in this question?

2. May 3, 2012

### HallsofIvy

Staff Emeritus
The first thing you need to do is to find the general solution to the associated homogeneous equation which is y''''- 4y''= 0. That has characteristic equation $r^4- r^2= r^2(r- 2)(r+ 2)= 0$ which has 2 and -2 as roots and 0 as a double root. The general solution to the associated homogeneous equation is $C_1x+ C_2+ C_3e^{2x}+ C_4e^{-2x}$.

And that should tell you that what you give cannot give anything but 0. Instead, you need to try something of the form $x^2(Ax^2+ Bx+ C)+ Dxe^{2x}$.

3. May 3, 2012

### physicsjock

thanks hallsofivy,

yea i used that general solution to apply variation of parameters,

well i used y1=e^(2x), y2 = e^(-2x) y3=x, y4 =1,

Ill do what you said and try somthing of the form you said,

I'm just not sure what this means

"And that should tell you that what you give cannot give anything but 0"

How did you know to multiply the polynomial by x^2 and the exponential by x in the particular solution? I kinda get why you pick x^2, since without it the polynomial is useless when you take the fourth derivative, but i wouldn't of been able to guess the x by the exponential,

Is there like a "check list" like there is for 2nd degree equations?