# Nonhomo with natural log

## Main Question or Discussion Point

y''-y'-30y=ln(t)

My attempt:
i tried to use the method of undetermined coefficients.
y''-y'-30y=ln(t)

Y(t)=A lnt
Y'(t)=A/t
Y''(t)= -A/t^2

I also tried:

Y(t) = A ln(t) + B 1/t + C 1/t^2

now im stuck....any help??

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dextercioby
Homework Helper
Can you at least solve the homogenous equation ?

dextercioby
Homework Helper
Maple returns this solution:

$$y^{\prime \prime }-y^{\prime }-30y=\ln x$$,

Exact solution is : $$y\left( x\right) =-\frac{1}{30}\ln x-\frac{1}{55}e^{-5x}\mbox{Ei}\left( 1,-5x\right) -\allowbreak \frac{1}{66}e^{6x}\mbox{Ei}\left( 1,6x\right) +C_{1}e^{-5x}+C_{2}e^{6x}$$

HallsofIvy