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NON-HOMOGENOUS
SECOND ORDER DE
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willbeuploadedon25th Jan2011
, .
PM
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- nd
complementary solution from its associated homogenousDE and then
find aparticular solutionbased on the rightside term.
Then, a particular solution of non-homogenous second
order DE can be obtained by:
1. Undetermined coefficien method rovided that theright side is
of common type).
. .
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Nonhomo enous2nd
orderDEA second order non-homogenous DE has the form:
1
With r(t)0
r t
The associated homogeneous differential equation to (1) is:
(2)
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To prove that Y1(t) - Y2(t) is a solution to (2) all we need to
do
Proof
s p ug t s nto t e erent a equat on an c ec t.
r t-r t=0
YP(t) is any solution to (1), then using the second part of
our
theorem we know that:
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Solution to non-homogenous 2nd order DE
The general solution to a 2nd order non-homogenous DE can
then
be written as:
yc t s ca e as comp mentary so ut on an p t) s ca e as
particular solution.
There are two common methods for findin articular solutions:
So, to solve a non-homogeneous DE, we will need to solve
theassociated homogeneous DE (2).
Undetermined Coefficients and Variation of Parameters. Both
have their advantages and disadvantages.
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Solution
by
Undetermined
CoefficientOne of the main advantages of this method is that it
reduces
the problem down to an algebra problem.
The method is quite simple: All that we need to do is look at
r(t) and make a guess to the
orm o YP(t) eav ng t e coe c ents un eterm ne .
Plug the guess YP(t) into the DE.
There are disadvantages to this method.
First, it will only work for a common class ofr(t). However,
ere are many unc ons ou ere or w c un e erm necoefficients
simply wont work.
Second, it is generally only useful for constant coefficient
DE.
Third, sometimes the solution may not be obtained by the
guess provided in the table.
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Summary of guess forY (t)
r(t)
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Example 1:
Find the general solution for the following non-homogenous
DE:
Solution:
teyyy
5312'4'' =
, .
corresponding homogenous DE is:012'4'' = yyy
From our previous lecture, we know the solution. The solution
will be thecomplimentary solution as:
tt
c ececty6
2
2
1)( +=
, p , .
Since :t
p
guesstAetYetg 55 )()( = = r(t)
Thus:t
p
t
p AeYAeY5''5' 255 ==
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xamp e cont .
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Example 2:
SolutionWe can make guess from the table:
Differentiate and plug into the DE,
The particular solution is:
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Example 3: Guess Yp(t) from r(t)
p -homogenous DE with the respective r(t):
1. ttetr 4)( = pp ettett +=+=
2. )10sin(16)( 7 tetr t=
23.
2
4. =
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SolutionbyVariationofParameters
Consider the differential equation:
)()(')('' trtqytpy =++
Assume that y1(t) and y2(t) are fundamental set of solutionsfor
the associated homogenous DE:
0)(')('' =++ tqytpy
Then a particular solution to the non-homogenous DE is:
dtyyW
tryydt
yyW
tryyYp += ),(
)(
),(
)(
21
12
21
21
Notice that Wis Wronskian of y1, y2.
n y1, y2 we ge rom e comp emen ary so u on o e correspon ng
homogenous DE:
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Wronskian is written as:
o u on y ar a ono arame ers
''21 yyW =
Example 4:
FindthegeneralsolutionforthefollowingDE:
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Exercise:
1.Findthegeneralsolutionto: tyyy 2sin35'2'' =++
2.Obtainthegeneralsolutionofthisnonhomogeneousdifferential
equationusingvariationofparametersmethod(Wronskian):x
exyyy2)1(4'4'' +=+
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