Take the homogeneous part and find the roots to the characteristic equation:
This means the characteristic solution is
.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With
and
, you're looking for a particular solution of the form
. The functions
satisfy
where
is the Wronskian determinant of the two characteristic solutions.
So you have
So you end up with a solution
but since
is already accounted for in the characteristic solution, the particular solution is then
so that the general solution is
A polynomial with a degree of 2 can be:
x² + 1
y² + 2
etc..
hope this helps
Answer:
<em>x</em> = -33.462
Step-by-step explanation:
-8.58/<em>x</em> = 3.9
<em>x</em> = 3.9 x -8.58
<em>x</em> = -33.462
Answer:
Step-by-step explanation: