# Nonhomo with natural log

1. Mar 20, 2008

### asourpatchkid

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??

2. Mar 20, 2008

### dextercioby

Can you at least solve the homogenous equation ?

3. Mar 20, 2008

### dextercioby

Maple returns this solution:

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

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

4. Mar 20, 2008

### HallsofIvy

Staff Emeritus
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.)

5. Mar 21, 2008

### asourpatchkid

all right, i just used the integral. thanks alot!