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??
 

Answers and Replies

dextercioby
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Can you at least solve the homogenous equation ?
 
dextercioby
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Maple returns this solution:

[tex] y^{\prime \prime }-y^{\prime }-30y=\ln x [/tex],

Exact solution is : [tex]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} [/tex]
 
HallsofIvy
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You can only use "undetermined coefficients" on equations where the right hand side is one of the functions we get as solutions to linear equations with constant coefficients- exponentials, polynomials, sine and cosine, and combinations of those. For other "right hand sides", you will have to use "variation of parameters".

(And, typically, you have to leave the solution in terms of an integral since that usually results in integrals that have no elementary formula.)
 
all right, i just used the integral. thanks alot!
 

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