Undetermined Coef. for Non-Homogeneous Cauchy-Euler Eq.

[cockroach]

For the non-homogeneous Cauchy-Euler Equation,

y''-(5/x)y'+(9/x^2)y=4x

I tried performing the substitution z=lnx and x=e^z. That transforms the equation into y''-6y'+9y=4e^z with respect to z. After finding the homogeneous solutions y=e^(3z) and y=ze^(3z), I do the method of undetermined coefficients with the yp=Ae^(z) as the particular solution. Substituted into y''-6y'+9y=4ez, the solution is A=1. So yp=e^z, which transforms to yp=x. But if I just use the homogeneous solutions y=e^(3z) and y=ze^(3z), convert them to y1=x^3 and y2=(x^3)lnx, and then do variation of parameters with respect to x, the solution is yp=2(x^3)ln^2(x) which transforms into yp=2e^(3z)z^2 as the particular solution. I don't know why the results don't match and what mistake I made? Is the undetermined coefficients method a legitimate way to solve these types of equations?

 

[galactus]

Regrettably, undetermined coefficients is limited to nonhomogeneous linear equations with constant coefficients.

Such as $\displaystyle a_{n}y^{(n)}+a_{n-1}y^{(n-1)}+\cdot\cdot\cdot a_{1}y'+a_{0}y=g(x)$.

In other words, the x's are giving you the fit. This is more like a Cauchy-Euler equation.

If we make the sub $\displaystyle y=x^{m}$, then we find the auxiliary equation is $\displaystyle m(m-1)-5m+9-0$

Which is $\displaystyle (m-3)^{2}$.

When there are repeated roots, then we have

$\displaystyle y_{c}=C_{1}x^{3}+C_{2}x^{3}ln(x)$

And $\displaystyle C_{1}x^{3}+C_{2}x^{3}ln(x)+2x^{3}ln^{2}(x)$

Your variation of parameters appears to be correct. Go with that.