70% is equal to 175$ dollars how many dollars are in 100%
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First note that 
is a regular singular point; in particular is a pole of order 1 for 
.
We seek a solution of the form
where
is to be determined. Differentiating, we have
and substituting into the ODE gives
The indicial polynomial,
, has roots at 
and 
. Because these roots are separated by an integer, we have to be a bit more careful, but we'll get back to this later.
When, we have the recurrence
valid for
. When 
, with 
, we find
and so on, with a general pattern of
Similarly, when
for , we find
and so on, with the general pattern
So the first indicial root admits the solution
which you can recognize as the power series for
and 
.
To be more precise, the second series actually converges to
, which doesn't satisfy the ODE. However, remember that the indicial equation had two roots that differed by a constant. When , we may seek a second solution of the form
where
. Substituting this into the ODE, you'll find that 
, and so we're left with
Expanding
, you'll see that all the terms 
with 
in the expansion of this new solutions are already accounted for, so this new solution really only adds one fundamental solution of the form 
. Adding this to , we end up with just 
.
This means the general solution for the ODE is
We seek a solution of the form
where
and substituting into the ODE gives
The indicial polynomial,
When
valid for
and so on, with a general pattern of
Similarly, when
and so on, with the general pattern
So the first indicial root admits the solution
which you can recognize as the power series for
To be more precise, the second series actually converges to
where
Expanding
This means the general solution for the ODE is