Particular Solutions of NonHomogeneous Linear Differential Equations with constant coefficients

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Particular Solutions of Non-Homogeneous Linear
Differential Equations with constant coefficients
Method of Undetermined Coefficients
In this lecture we discuss the Method of
undetermined Coefficients.
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Consider the second order non-homogeneous linear
differential equation with constant coefficients
Here p and q are constants.
We discuss the particular solutions for different
RHS h(x).
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Case (i)
a constant
Thus we want to find a function y of x satisfying
the d.e.
We ask what sort of function y will be.
Obviously a first guess is that it should be
a multiple of
So we assume a particular solution as
a constant.
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Substituting y A eax in the above d.e., we get
Thus
provided the Denominator
Hence a particular solution is
with A as found above.
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Example 1
Find a particular solution of
Solution Let a particular solution be
Substituting, we get
Hence A 1/5
And so a particular solution is
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What happens if
This means a is a root of the auxiliary equation
Or
is a solution of the associated
homogeneous l.d.e.
is a part of the Complementary
We say
function.
It is suggested that we take a particular
solution as
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where r is the least positive integer such that
is NOT part of the complementary
function. Substituting for y, Dy, D2y we can
find A.
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Example 2
Find a particular solution of
Solution Now the auxiliary equation is
Roots are 2, 3
Hence the Complementary function is
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Since
is a part of the C.F. but
is not, we take a particular solution as
6?
-5?
1?
Adding we get
or
Thus we get
A 2
And so a particular solution is
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Case (ii)
k1, k2 constants
Thus we want to find a function y of x satisfying
the d.e.
We assume a particular solution as
We substitute for y, Dy, D2y and equating the
coefficients of cos bx and sin bx, find the
constants A1 and A2 (if possible).
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Example 3
Find a particular solution of
Solution Let a particular solution be
-1
1
1
Adding we get
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gives
Solving we get
Hence a particular solution is
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where c1 and c2 are arbitrary constants
If
is a solution of the associated
homogeneous l.d.e.
We say
is a part of the Complementary
function.
It is suggested that we take a particular
solution as
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as our trial solution
We substitute for y, Dy, D2y and equating the
coefficients of cos bx and sin bx, find the
constants A1 and A2 (if possible).
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Example 4
Find a particular solution of
Solution Let a particular solution be
1
0
1
Adding we get
So we are not able to find A1, A2.
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We observe that (RHS) function cos x is a part of
the C.F. So we assume a particular solution as
1
0
1
Adding we get
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Thus A1 0, A2 3/2
Hence a particular solution is
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Case(iii) h(x) is a polynomial of degree n. We
assume a particular solution as
Substituting for y, Dy, .. and equating the
coefficients of like powers of x on both sides,
we get As and hence a particular solution.
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Example 5
Find a particular solution of
Solution Let a particular solution be
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-6
1
1
Adding we get
gives
Hence a particular solution is
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If the constant q happens to be zero, then this
procedure gives xn-1 as the highest power of x on
the left of (), so in this case we take our
trial solution in the form
If p and q are both zero, then () can be solved
at once by direct integration.
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Example 6
Find a particular solution of
Solution Let a particular solution be
0
3
1
Adding we get
is not meaningful and does not give A0.
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Note C.F. is
We observe this is due to the fact that y 1 is
a part of the Complementary function. Hence a
particular solution is taken as
0
3
1
Adding we get
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gives
Solving, we get
Hence a particular solution is
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Example 6
Find a particular solution of
Solution Let a particular solution be
10
6
1
Adding we get
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gives
or
or
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or
or
Thus a particular solution is
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Principle of Superposition
Consider the linear d.e.
where
Suppose
And that
is a particular solution of
is a particular solution of
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Then
is a particular solution of
This is known as the principle of superposition.
Example 7 Find the general solution of the d.e.
Solution The general solution is
C.F. Particular Solution
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Step 1 To find yh, the C.F.
Auxiliary Equation
Roots m
3, -2
Hence the Complementary function is
c1, c2 arbitrary constants
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Step 2 To find yp, a particular solution
Let
be a particular solution of the d.e.
Let
be a particular solution of the d.e.
Let
be a particular solution of the d.e.
Let
be a particular solution of the d.e.
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Then by the principle of superposition, a
particular solution of the given d.e. is
Consider
is a part of the C.F. but
is not,
Since
we assume a particular solution as
Substituting for y, Dy, D2y, we get
Hence
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Consider
We assume a particular solution as
Substituting for y, Dy, D2y, we get
Hence
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Consider
We assume a particular solution as
Substituting for y, Dy, D2y, we get
Hence
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Consider
We assume a particular solution as
Substituting for y, Dy, D2y, we get
Hence
Hence
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We assume a particular solution of
as
We assume a particular solution of
as
We also use multiply by the least power of x
rule in case any term of the Given RHS is a part
of the C.F.
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Example 8 Find the general solution of the d.e.
Solution We first find the complementary function.
Auxiliary Equation
Roots m
Hence the Complementary function is
c1, c2 arbitrary constants
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Now we find the particular solution.
As
is a part of the C.F., we take a
particular solution as
2
-2
1
Adding, we get
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gives
or
Hence a particular solution is
Hence the general solution is
i.e.
c1, c2 arbitrary constants