Can't solve this higher order problem

[CDS]

y'''-6y''+8y'=ln x

i can find yc, but i can't find the yp part

i used the Wronskian method and i plug the values to find u1, u2, u3
but i'm stuck withe integration part

can't solve this integral

S(lnx)(e^-4x) dx

please help

anyone has any idea how can i work with this

 

[royhaas]

Check the digamma function for the integral.

 

[galactus]

The integral you have is not doable by elementary methods. But, it is expressible in terms of the Psi function, or as Roy said, the Digamma function.

Represented by \(\displaystyle {\psi}\)

\(\displaystyle \int ln(x)e^{-4x}dx\)

Make the sub \(\displaystyle u=4x, \;\ du=4dx\)

and then we can express it as \(\displaystyle \frac{1}{4}\int\left(ln(u)-ln(4)\right]e^{-u}du\)

But \(\displaystyle {\Psi}(x)=\frac{{\Gamma}'(x)}{{\Gamma}(x)}\)

The Digamma function is the logarithmic derivative of the Gamma function.

Note that \(\displaystyle \frac{d}{dp}{\Gamma}(p)=\int_{0}^{\infty}x^{p-1}e^{-x}ln(x)dx\)

and \(\displaystyle {\Gamma}(p)=\int_{0}^{\infty}x^{p-1}e^{-x}dx\)

Can you see the relationship with your integral?.

 

[galactus]

\(\displaystyle y'''-6y''+8y'=ln(x)\)

You can use VOP on higher order DE's as well by setting up the Wronskians.

\(\displaystyle m^{3}-6m^{2}+8m=0\)

\(\displaystyle m=0, \;\ 2, \;\ 4\)

\(\displaystyle y_{c}=C_{1}+C_{2}e^{2x}+C_{3}e^{4x}\)


\(\displaystyle W=\begin{vmatrix}1&e^{2x}&e^{4x}\\0&2e^{2x}&4e^{4x}\\0&4e^{2x}&16e^{4x}\end{vmatrix}\)

Can you continue?.