Let the vector v have an initial point at (3,−2) and a terminal point at (3,−7). Determine the components of vector v
2 answers:
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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
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
With
where
So you have
So you end up with a solution
but since
so that the general solution is
Midpoint formula
mid of (x1,y1) and (x2,y2) is
so
(22,15)=
therefor
22=
and
15=
slv each
22=
times both sides by 2
44=18+x1
minus 18 both sides
26=x1
15=
times both sides by 2
30=6+y1
minus 6
24=y1
the other point is (26,24)
mid of (x1,y1) and (x2,y2) is
so
(22,15)=
therefor
22=
and
15=
slv each
22=
times both sides by 2
44=18+x1
minus 18 both sides
26=x1
15=
times both sides by 2
30=6+y1
minus 6
24=y1
the other point is (26,24)