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1. SecondOrder Linear NonHomogeneous Differential Equations With

In this section, we'll use the abbreviations:
“DE” 
for 
“differential equation”, 
“GS” 
for 
“general solution”, 
“HDE” 
for 
“homogeneous differential equation”, 
“HE” 
for 
“homogeneous equation”, 
“NHDE” 
for 
“nonhomogeneous differential equation”, 
“NHE” 
for 
“nonhomogeneous equation”, and 
“PS” 
for 
“particular solution”. 
Definition 1.1 – SecondOrder Linear NonHomogeneous
Differential Equations With
Constant
Coefficients
A differential equation of the form: y'' + by' + cy = f (x), where b and c are constants and f
is a continuous function that isn't identically 0, is called a secondorder
linear 
Note that there are only the terms for y,
y', and y''
on the lefthand side. Everything else must be on the righthand side and is
a part of f. Distinguish between the functions y and f. Observe
that while the equation is linear, each function y,
y', y'',
and
f (x) doesn't have to be linear.
For example, y = 2x + 3 is
linear while y = x^{2} + 3 and y = e^{2}^{x}
+ 3 aren't.
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2. General Solutions Of The NonHomogeneous Equations 
Suppose y_{n}_{1} is a PS of the NHE:
y'' + by' + cy = f (x)
and y_{h} is the GS of the corresponding HE:
y'' + by' + cy = 0.
Then intuitively the sum y_{h}
+ y_{n}_{1} should be the
GS of the NHE, because: (1) when we substitute y_{h}
+ y_{n}_{1} into the NHE,
the
expression containing y_{h} and its
derivatives equals 0, since y_{h}
is the GS of the HE, (2) the expression containing y_{n}_{1} and its
derivatives equals f (x),
since y_{n}_{1} is a PS of the
NHE, and (3) there are 2 arbitrary constants in y_{h}
+ y_{n}_{1}, they're the
arbitrary
constants of y_{h}. And so the
substituted sum equals 0 + f (x)
= f (x). This
intuition is confirmed true in the theorem below.
Theorem 2.1 – General Solutions Of The NonHomogeneous Equations
Let: ((GS Of NHE) = (GS Of HE) + (A PS Of NHE)). 
Proof
First let's prove that y_{h}
+ y_{n}_{1} is a solution
of the NHE [2.1]. We have:
Thus y_{n}_{2} – y_{n}_{1} is a PS of the
HE. Let y_{h}_{1} be that PS, so
that y_{n}_{2} – y_{n}_{1} = y_{h}_{1}, or y_{n}_{2} = y_{h}_{1} + y_{n}_{1}. Now y_{h}_{1} can be
obtained from y_{h}.
As a consequence, y_{n}_{2} can be
obtained from y_{h} + y_{n}_{1}. Hence y_{h} + y_{n}_{1} is the GS of
the NHE [2.1].
EOP
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3. Principle Of Superposition 
Because 0 + f (x) = f (x), by Theorem 2.1 we see that:
(GS of y'' + by' + cy = 0 + f (x)) = (GS of y'' + by' + cy = 0) + (PS of y'' + by' + cy = f (x)).
This leads us to investigate to see if the property:
(solution of y'' + by' + cy = f_{1} (x) + f_{2} (x)) = (solution of y'' + by' + cy = f_{1} (x)) + (solution of y'' + by' + cy = f_{2} (x))
is also true for any functions f_{1} (x)
and f_{2} (x).
Intuitively this is obviously true. Let y_{1} be a solution
of y'' + by' + cy = f_{1} (x) and
y_{2} a solution of y''
+ by' + cy = f_{2} (x).
Substituting y_{1} + y_{2} for y in the expression y''
+ by' + cy, we get f_{1} (x)
+ f_{2} (x),
because
the expression containing y_{1} and its
derivatives equals f_{1} (x)
and the expression containing y_{2} and its
derivatives equals f_{2} (x).
It turns out that this property is indeed true, as asserted by
Theorem 3.1 below, which establishes the relationship between the
solutions of the equations:
y''
+ by' + cy = f _{1}(x),
y'' + by'
+ cy = f _{2}(x), and
y'' + by'
+ cy = f _{1}(x) + f _{2}(x),
the relationship being that the sum of a solution of the 1st
equation and a solution of the 2nd equation is a solution of the 3rd
equation. Letting “RHS” stands for “righthand side”, this property can be
thought of as:
(solution when RHS is) ( f_{1}(x) + f_{2}(x)) = (solution when RHS is) ( f_{1}(x)) + (solution when RHS is) ( f_{2}(x)).
This linearity property of the solutions is called the principle
of superposition, as a solution of the 3rd equation is obtained by
superimposing a solution of 1st equation on a solution of the 2nd equation.
Theorem 3.1 – Principle
Of Superposition
If y_{1} is a solution of the equation: This property is called the principle
of superposition, because a solution of Eq. [3.3] is obtained by
superimposing a 
Proof
So y_{1} + y_{2} is a solution of y'' + by' + cy = f_{1}(x) + f_{2}(x).
EOP
a. The sum of a solution of Eq. [3.1] and a solution
of Eq. [3.2] is a solution of Eq. [3.3], not a solution of the equation that's
the sum of Eqs. [3.1] and [3.2],
which is 2( y'' + by'
+ cy) = f_{1}(x) + f_{2}(x), or y''
+ by' + cy = (1/2)( f_{1}(x) + f_{2}(x)).
b. It's straightforward to show that the principle of
superposition is valid for any number of equations. Let n
be any integer
greater than or equal to 2. If y_{1}, y_{2}, ..., and y_{n} are the solutions of
the equations y'' + by' + cy = f_{1}(x), y''
+ by' + cy =
f_{2}(x), ..., and y''
+ by' + cy = f_{n}(x) respectively, then y_{1} + y_{2} + ... + y_{n}
is a solution of the equation y''
+ by' + cy = f_{1}(x) +
f_{2}(x) + ... + f_{n}(x).
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4. The Method Of Undetermined Coefficients 
Clearly Theorem 2.1 tells us that
to solve (find the GS of) a secondorder linear NHDE with constant
coefficients, we have to
find the GS of the corresponding HE and a PS of the NHE, then we add them up to
get the GS of the NHDE. We already know
how to find the GS of a HE. We'll present a method to determine a PS of the
NHE, called the method of undetermined
coefficients. The reason for this naming is clear later on. This method
applies only to some types of functions on the righthand
side of the NHE, ie the function f (x).
When f (x) Is A Non0 Constant
First we show that when f (x)
is constant, we can find a PS of the NHE simply by inspection. Suppose we're to
find a PS of the
NHE:
y'' – 7y' + 10y = 20.
Since a non0 constant, 20 in this case, is a polynomial (of
degree 0), finding a PS of the NHE where f (x)
is a polynomial can
be done by the method of undetermined coefficients, as illustrated in Example 4.2 below.
Finding a PS of the NHE where f (x)
is a non0 constant can also be done by inspection. The function f (x) is the constant 20. If
y is a constant say k,
then y' = 0 and y'' = 0, so when we compute y'' – 7y'
+ 10y, it's a constant, namely 10k, and thus we
can choose k so that y''
– 7y' + 10y = 20. We'll then have 10k
= 20, consequently k = 20/10 =
2. Hence we'll try the function
y = 2 as a PS of the NHE. We use inspection in
Example 4.1 below.
Find the GS of the equation y'' – 7y' + 10y = 20. Check the answer.
Solution
For a PS of the given NHE y''
– 7y' + 10y = 20, let's try y
= 20/10 = 2. Then y' = 0, y'' = 0, and y''
– 7y' + 10y = 0 – 7(0)
+ 10(2) = 20. So a PS of the NHE is y = 2.
Thus the GS of the NHE is y = c_{1}e^{5}^{x} + c_{2}e^{2}^{x} + 2.
Check:
y = c_{1}e^{5}^{x}
+ c_{2}e^{2}^{x}
+ 2,
y'
= 5c_{1}e^{5}^{x}
+ 2c_{2}e^{2}^{x},
y''
= 25c_{1}e^{5}^{x}
+ 4c_{2}e^{2}^{x},
y'' – 7y'
+ 10y = (25c_{1}e^{5}^{x} + 4c_{2}e^{2}^{x})
– 7(5c_{1}e^{5}^{x}
+ 2c_{2}e^{2}^{x})
+ 10(c_{1}e^{5}^{x}
+ c_{2}e^{2}^{x}
+ 2) = 20.
Indeed the GS y = c_{1}e^{5}^{x}
+ c_{2}e^{2}^{x}
+ 2 is correct.
EOS
When f (x) Is A Polynomial Function
Let's find a PS of the NHE:
y'' + 2y' – 3y = 2x.
The function f (x)
is a linear function, f (x)
= 2x. So y
should also be a linear function y = a_{1}x + a_{0}, so that when we compute
y'' + 2y'
– 3y, it's also a linear function, and
thus we can select the coefficients a_{1} and a_{0} that make y''
+ 2y' – 3y equal 2x.
Let's try y = ax,
where a is a constant to be determined such
that y'' + 2y' – 3y = 2x. We have:
y = ax, y' = a, y''
= 0,
2x = y''
+ 2y' – 3y = 0 + 2a – 3ax = – 3ax + 2a,
– 3ax + 2a
= 2x,
these 2 linear functions are the same thing if the coefficients of like powers
of x are equal; so let's equate the
coefficients of
like powers of x:
a = –2/3 and a = 0, impossible,
the system has no solutions.
The trial PS y = ax doesn't work.
So what do we do? Ok, there's one remaining thing to do:
it's to try the linear function of the general form y
= a_{1}x + a_{0}, where
a_{1} and a_{0} are constants
to be determined such that y''
+ 2y' – 3y = 2x. We have:
y = a_{1}x + a_{0}, y' = a_{1}, y'' = 0,
2x = y'' + 2y'
– 3y = 0 + 2a_{1} – 3(a_{1}x + a_{0}) = – 3a_{1}x + 2a_{1} – 3a_{0},
– 3a_{1}x + 2a_{1} – 3a_{0} = 2x,
these 2 linear functions are the same thing if the coefficients of like powers
of x are equal; so let's equate the
coefficients of
like powers of x:
solving this system we get a_{1} = –2/3 and a_{0} = 2(–2/3)/3 = –4/9.
The trial PS y = a_{1}x + a_{0} works. A PS of the NHE is y = – (2/3)x – 4/9.
Solve the NHE y''
+ 2y' – 3y = 2x by going
thru the following steps:
a. Solve the HE y''
+ 2y' – 3y = 0.
b. Find a PS of the given NHE. Verify the answer.
c. Form the GS of the NHE. Verify the answer.
Solution
Verification:
Indeed the GS is correct.
EOS
In Example
4.2 part b, since 2x is a linear
function, a PS y should also be a linear
function, so that y' and y'' also are, then y''
+ 2y' – 3y also
is, thus we can make it equal the linear function 2x.
Consequently, as a PS, we try the linear function of the
general form y = a_{1}x + a_{0}, not of the
nongeneral form y = ax,
as seen in the discussion preceding Example 4.2. We compute
y', y'',
then y'' + 2y' – 3y. For y'' + 2y'
– 3y to equal 2x,
its xexpression must equal 2x. Hence we
equate its xexpression
with 2x. For the 2 expressions to be the same thing, ie to be equal for all x, the coefficients of like powers of x must be equal.
It follows that we equate the coefficients of like powers of x (the constants are the coefficients of x^{0} = 1), get a system of 2
equations in a_{1} and a_{0}, solve it,
obtain a_{1} = –2/3 and a_{0} = –4/9.
Therefore the PS is y = – (2/3)x – 4/9.
We try the trial solution y
= a_{1}x + a_{0}, where the coefficients a_{1} and a_{0} are to be determined. Of course initially they're
undetermined. Hence this method of trying a PS with initiallyundetermined
coefficients is called the method of
undetermined coefficients.
As asked, we verify the PS and the GS of the NHE, and find
that indeed they're correct. The procedure to determine a PS
assures that the PS must be correct, and thus, by Theorem
2.1, the GS must also be correct. Example
4.2 asks to verify simply
because at the first contact with the method, it's good to assure you clearly
that of course the method works. In solutions to
problems, if you're not asked to verify, then you don't have to verify.
Solve y'' – y = x^{2}.
The function f (x)
is a quadratic function, f(x) = x^{2}. So y should also be a quadratic function, so that
when we compute y'' – y,
it's also a quadratic function and thus has a chance to equal x^{2}. This convinces us, in using
the method of undetermined
coefficients, to try as a PS of the given NHE the quadratic function of the
general form y = a_{2}x^{2} + a_{1}x + a_{0}, where a_{2}, a_{1}, and a_{0}
are constants to be determined in such a way that y''
– y = x^{2}. As in the
case for f (x) being a
linear function, we must try the
general form of the polynomial, in this case the general form of the quadratic
function.
Solution
a_{2} = –1, a_{1} = 0, a_{0} = 2a_{2} = 2(–1) = –2,
y = – x^{2} – 2.
The GE of the NHE is y = c_{1}e^{x}
+ c_{2}e^{–}^{x}
– x^{2} – 2.
EOS
When f (x) Is An Exponential Function
Let's find the working trial solution of the NHE:
y'' – 3y' + y = 7e^{2}^{x}.
The function f (x)
is an exponential function with
exponent 2x, f (x)
= 7e^{2}^{x}. So y should also be an exponential function with
the same exponent 2x, so that
when we compute y'' – 3y' + y, it's also
an exponential function with
the same exponent 2x, as
the derivative of any order of e^{2}^{x}
is a constant multiple of e^{2}^{x},
and thus y'' – 3y' + y can be made
to equal 7e^{2}^{x}. The trial
solution is the general form of the exponential
function e^{2}^{x}, which is y = ae^{2}^{x}.
This trial solution works, as shown in
Example
4.4 below. Note that the trial solution when f (x) = ke^{sx}, where k and s are constants, is ae^{sx}, not ake^{sx}.
Solve the DE y'' – 3y' + y = 7e^{2}^{x}.
Solution
EOS
When f (x) Is A Sine Or Cosine Function
Let's determine the working trial solution of the NHE:
y'' + 5y' + 4y = 2 sin x.
The function f (x)
is a sine function, f (x)
= 2 sin x. So the xexpression of y''
+ 5y' + 4y must be able to equal 2 sin x. This
suggests that we should use a sine function as a trial solution. The derivative
of any order of sin x is a
constant multiple of
either sin x or cos
x. This reenforces the suggestion to use a sine
function. Thus let's try y = a sin x, where a is a constant to
be determined so that y''
+ 5y' + 4y = 2 sin x.
We have:
y = a sin x, y'
= a cos x, y'' = – a sin
x,
2 sin x = y'' + 5y'
+ 4y = – a
sin x + 5a cos x + 4a sin x = 3a sin x + 5a cos x,
3a sin x + 5a cos
x = 2 sin x,
equate the coefficients of like trigonometric functions of x:
this system has no solutions.
So the trial solution y = a sin x doesn't work. The reason is discussed after the Solution of Example 4.5 below.
So what do we do? Ok, there's one remaining thing to do:
it's to try the cosine+sine function of the general form y
= A cos
x +
B sin x, where A
and B are
constants to be determined so that y''
+ 5y' + 4y = 2 sin x.
This trial PS works, as seen in
Example 4.5 below. The reason is discussed after the Solution of Example 4.5.
Solve y'' + 5y' + 4y = 2 sin x.
Solution
EOS
For the trial solution y = ax, the xexpression
of y'' + 5y' + 4y is 3a sin x + 5a cos x, where the coefficients of cos x and
sin x are products containing the same factor a. When we remove the term 5a cos
x by setting a = 0, the term 3a sin
x will
also disappear. That's the reason why y = a sin
x doesn't work.
For the trial solution y = A cos x + B
sin x, the xexpression of y''
+ 5y' + 4y is (– 5A
+ 3B) sin
x + (3A
+ 5B) cos
x, where
the coefficients of cos x and sin x aren't products containing the same factor that has to be set to 0.
When we remove the
term (3A +
5B) cos
x by
setting 3A +
5B = 0, the term (– 5A + 3B)
sin x won't disappear. That's the reason why y = A cos
x
+ B sin x works.
Find the general solution of the DE y'' – 3y' + 2y = 2 cos 4x.
Similar to the discussions surrounding Example 4.5. The general form of the
cosine+sine function cos 4x + sin
4x is A cos 4x
+ B sin
4x.
Solution
EOS
Eligible Functions
In this section, eligible functions refer to functions that
are eligible for the method of undetermined coefficients. They're
represented by the function f (x)
in the NHE:
y''
+ by' + cy = f (x).
They're the polynomial, sine, cosine, and exponential functions.
If f (x)
is a polynomial function of degree n: derivatives
of orders 0 thru to 2, which appear on the lefthand side of the
equation, of the polynomial function f (x)
of degree n are polynomial functions of degree n and less; so the trial PS of the NHE
is a polynomial function of degree n of the
general form:
y = a_{n}x^{n} + a_{n}_{–1}x^{n}^{–1} + a_{n}_{–2}x^{n}^{–2} + ... + a_{2}x^{2} + a_{1}x + a_{0},
so that the two sides can be made equal by equating the coefficients of like powers of x.
If f (x)
is a sine or cosine function: derivatives of the sine function f (x) = r
sin sx and of the
cosine function f (x) = r cos sx
are each constant multiples of the sine and cosine functions sin
sx and cos sx;
so the trial PS is the cosine+sine function of the
general form:
y = A cos sx + B sin sx.
If f (x)
is an exponential function: derivatives of the exponential function f (x) = re^{sx}
are constant multiples of the exponential
function e^{sx}; so the trial PS is
the exponential of the general form:
y = ae^{sx}.
For noneligible functions, consider the following example:
If f (x)
= ln x, letting the trial solution
be y = a
ln x would lead to y'' + by'
+ cy containing a constant multiple of
–1/x^{2}, and
thus y'' + by' + cy can't be
made equal to ln x.
There are situations where the trial PSs have to be
modified. This modification is discussed in the next section, where part 2 of
this topic is presented. Thus if you're encountering this topic for the first
time, you're reminded that your knowledge of it will be
complete only after you successfully finish its part 2.
When f (x) Is A Sum Of Eligible Functions
Consider the equation y''
– y = x^{2} + 3 cos
x.
a. Solve the equation y''
– y = x^{2}.
b. Solve the equation y''
– y = 3 cos x.
c. Solve the equation y''
– y = x^{2} + 3 cos
x using the principle of superposition.
d. Solve the equation y''
– y = x^{2} + 3 cos
x using the method of undetermined coefficients.
For part d. The function f (x)
is the sum of a quadratic function and a cosine function, f (x)
= x^{2} + 3 cos x. So naturally, as a
PS of the NHE y'' – y = x^{2} + 3 cos
x, we'll try the sum of the general quadratic
function and the general form of the
cosine+sine function, that sum being y = a_{2}x^{2} + a_{1}x + a_{0} + A cos x + B sin x, so that
when we compute y'' – y, it'll contain x^{2}
and cos x, and thus
it'll be possible to make it equal x^{2} + 3 cos
x by the process of equating coefficients.
Solution
a_{2} = –1, a_{1} = 0, a_{0} = 2(–1) = –2,
a PS of the NHE is y = – x^{2} – 2.
So the required GS is:
y = c_{1}e^{x} + c_{2}e^{–}^{x} – x^{2} – 2.
b. By part a, the GS of the HE y'' – y = 0 is y = c_{1}e^{x} + c_{2}e^{–}^{x}.
For a PS of the NHE y'' – y = 3 cos x, let y = A cos x + B sin x. We have:
y = A cos x + B sin x, y' = – A sin x + B cos x, y'' = – A cos x – B sin x,
3 cos
x = y''
– y = (– A cos x – B sin x) – (A
cos x + B sin x) = – 2A
cos x – 2B sin x,
– 2A cos x – 2B sin x = 3 cos x,
d. By part a, the GS of the HE y'' – y = 0 is y = c_{1}e^{x} + c_{2}e^{–}^{x}.
For a PS of the NHE y'' – y = x^{2} + 3 cos x, let y = a_{2}x^{2} + a_{1}x + a_{0} + A cos x + B sin x. We have:
y = a_{2}x^{2} + a_{1}x + a_{0} + A cos x + B sin x, y' = 2a_{2}x + a_{1} – A sin x + B cos x, y'' = 2a_{2} – A cos x – B sin x,
EOS
As Example 4.7 part c shows, when f (x)
is the sum of two or more eligible functions, we can use the principle of
superposition, as shown in Theorem 3.1. A PS of:
y'' + by' + cy = f_{1}(x) + f_{2}(x) + ... f_{n}(x)
equals the sum of PSs of:
y'' + by' + cy = f_{1}(x), y'' + by' + cy = f_{2}(x), ..., and y'' + by' + cy = f_{n}(x).
When f (x) Is A Product Of A Polynomial Function And An Exponential Function
Let's find a PS of the NHE:
y'' + y = xe^{2}^{x}.
The function f (x)
is the product of a polynomial function of degree 1 and an exponential function, f (x) = xe^{2}^{x}. So naturally, as
a PS of this equation, we'll try the product of the general polynomial function
of degree 1 and the general form of the
exponential function e^{2}^{x}, that product being y = (a_{1}x + a_{0})ae^{2}^{x}, so that when we
compute y'' + y, it'll contain the product xe^{2}^{x},
and thus it'll be possible to make it equal xe^{2}^{x}
by the process of equating coefficients. We have:
y = (a_{1}x + a_{0})ae^{2}^{x},
y' =
aa_{1}e^{2}^{x}
+ (2aa_{1}x + 2aa_{0})e^{2}^{x},
y''
= 2aa_{1}e^{2}^{x}
+ 2aa_{1}e^{2}^{x}
+ (4aa_{1}x + 4aa_{0})e^{2}^{x}
= 4aa_{1}e^{2}^{x}
+ (4aa_{1}x + 4aa_{0})e^{2}^{x},
xe^{2}^{x}
= y'' + y = 4aa_{1}e^{2}^{x}
+ (4aa_{1}x + 4aa_{0})e^{2}^{x}
+ (a_{1}x + a_{0})ae^{2}^{x}
= (5aa_{1}x + 4aa_{1} + 5aa_{0})e^{2}^{x},
(5aa_{1}x + 4aa_{1} + 5aa_{0})e^{2}^{x}
= xe^{2}^{x},
5aa_{1}x + 4aa_{1} + 5aa_{0} = x,
a_{1} = 1/(5a), a_{0} = – 4a(1/(5a)) / (5a) = – 4/(25a),
letting a = 1 we get a_{1} = 1/5 and a_{0} = –4/25, and a PS is y = ((1/5)x – (4/25))e^{2}^{x}.
Now (a_{1}x + a_{0})(ae^{2}^{x}) = (aa_{1}x + aa_{0})e^{2}^{x},
and aa_{1} and aa_{0} are constants.
Thus we can simply use y = (A_{1}x
+ A_{0})e^{2}^{x}, where A_{1}
and A_{0} are constants,
as the trial PS. Reutilizing the letters a_{1} and a_{0} in place of A_{1} and A_{0} respectively, our trial PS is y = (a_{1}x
+ a_{0})e^{2}^{x}.
This is the same as using a = 1. It
works all right, giving the same PS, as seen in Example 4.8 below. When a
constant can be combined with all the remaining constants, it can be removed.
Consider the equation y''
+ y = xe^{2}^{x}.
a. Solve the equation y''
+ y = x.
b. Solve the equation y''
+ y = e^{2}^{x}.
c. Solve the equation y''
+ y = xe^{2}^{x}.
d. Does the GS of y''
+ y = xe^{2}^{x} equal the product of
the GS of y'' + y = x and the GS
of y'' + y = e^{2}^{x}?
Solution
a_{1} = 1, a_{0} = 0,
a PS of the NHE is y = x.
So the GS of y'' + y = x is:
y = x + c_{1} cos x + c_{2} sin x.
b. By part a, the GS of the HE y'' + y = 0 is y = c_{1} cos x + c_{2} sin x.
For a PS of the NHE y'' + y = e^{2}^{x}, let y = ae^{2}^{x}. We have:
y = ae^{2}^{x}, y' = 2ae^{2}^{x}, y'' = 4ae^{2}^{x},
e^{2}^{x} = y''
+ y = 4ae^{2}^{x}
+ ae^{2}^{x} = 5ae^{2}^{x},
5ae^{2}^{x}
= e^{2}^{x},
5a = 1,
a = 1/5,
a PS of the NHE is y = (1/5)e^{2}^{x}.
So the GS of y'' + y = e^{2}^{x} is:
a_{1} =
1/5,
a_{0} = –
4(1/5)/5 = –4/25,
a PS of the NHE is y = ((1/5)x – (4/25))e^{2}^{x}.
So the GS of y'' + y = xe^{2}^{x}
is:
The product [1] x [2] contains the term x(c_{1} cos x) = c_{1}x cos x, which [3] doesn't contain. So the product [1]
x [2] can't
equal [3]. Thus the answer is no.
EOS
When f (x) Is A Product Of A Polynomial Function And A Sine Or Cosine Function
Let's determine the working trial PS of the NHDE:
y'' – y = x sin x.
Let's try y = (a_{1}x + a_{0})(A cos x + B sin x). We have:
y = (a_{1}x + a_{0})(A
cos x + B
sin x),
y'
= a_{1}(A
cos x + B
sin x) + (a_{1}x + a_{0})(– A sin x + B cos x),
y''
= a_{1}(– A
sin x + B
cos x) + a_{1}(– A sin x + B cos x) + (a_{1}x + a_{0})(– A
cos x – B
sin x)
= 2a_{1}(– A
sin x + B
cos x) – (a_{1}x + a_{0})(A cos x + B sin x),
x sin
x = y'' – y
= 2a_{1}(– A sin x + B cos x) – (a_{1}x + a_{0})(A
cos x + B
sin x) – (a_{1}x + a_{0})(A cos x + B sin x)
= 2a_{1}(– A sin x + B cos x) – 2(a_{1}x + a_{0})(A
cos x + B
sin x),
= (– 2a_{1}A – 2a_{1}xB
– 2a_{0}B)
sin x + (2a_{1}B – 2a_{1}xA
– 2a_{0}A)
cos x
= (– 2a_{1}Bx
– 2a_{1}A
– 2a_{0}B)
sin x + (– 2a_{1}xA
+ 2a_{1}B
– 2a_{0}A)
cos x
(– 2a_{1}Bx – 2a_{1}A – 2a_{0}B) sin x
+ (– 2a_{1}Ax + 2a_{1}B – 2a_{0}A) cos x
= x sin x,
as [1] and [3] show, the system has no solution, the trial solution y = (a_{1}x + a_{0})(cos
x + sin x)
fails too.
In Example 4.8, we have f (x)
= xe^{2}^{x}, and the working trial
solution is y = (a_{1}x + a_{0})e^{2}^{x}, obtained by
multiplying the coefficient
of e^{2}^{x} into the polynomial,
so that e^{2}^{x} has a coefficient of
1. Let's try the same strategy: move A
and B from A cos x + B
sin x
to the polynomial, so that cos x and sin
x each have a coefficient of 1. We have:
y = (a_{1}x + a_{0})(A cos x + B sin x) = (a_{1}x + a_{0})A cos x + (a_{1}x + a_{0}) B sin x = (a_{1}Ax + a_{0}A) cos x + (a_{1}Bx + a_{0}B) sin x.
Let's reuse the letters a_{1} and a_{0} for the products a_{1}A and a_{0}A respectively, and let b_{1} = a_{1}B and b_{0} = a_{0}B.
Then y =
(a_{1}x + a_{0}) cos x + (b_{1}x + b_{0}) sin x. That's
what we want! Let's try it: y = (a_{1}x + a_{0}) cos x + (b_{1}x + b_{0}) sin x, done in
Example 4.9 below. It works, as seen in the Solution! Wow!
Example 4.9
Solve the NHDE y''
– y = x
sin x.
For a PS of the given NHE y'' – y = x sin x, let y = (a_{1}x + a_{0}) cos x + (b_{1}x + b_{0}) sin x. We have:
y = (a_{1}x + a_{0}) cos
x + (b_{1}x + b_{0}) sin x,
y'
= a_{1} cos x – (a_{1}x + a_{0}) sin
x + b_{1} sin x + (b_{1}x + b_{0}) cos x,
y''
= – a_{1} sin x – a_{1} sin
x – (a_{1}x + a_{0}) cos x + b_{1} cos x + b_{1} cos x – (b_{1}x + b_{0}) sin x
= (– a_{1}x – a_{0} + 2b_{1}) cos
x + (– b_{1}x – 2a_{1} – b_{0}) sin
x,
x sin
x = y'' – y
= ((– a_{1}x – a_{0} + 2b_{1}) cos
x + (– b_{1}x – 2a_{1} – b_{0}) sin
x) – ((a_{1}x + a_{0}) cos x + (b_{1}x + b_{0}) sin x)
= (– 2b_{1}x – 2a_{1} – 2b_{0}) sin
x + (– 2a_{1}x – 2a_{0} + 2b_{1}) cos
x,
(– 2b_{1}x – 2a_{1} – 2b_{0}) sin x + (– 2a_{1}x – 2a_{0} + 2b_{1}) cos
x = x sin x,
b_{1} = –1/2, a_{1} = 0, b_{0} = 0, a_{0} = – 2(–1/2) /
(–2) = –1/2,
a PS of the NHE is y = – (1/2) cos x – (1/2)x sin x, or y = – (1/2)(cos x + x sin x).
Thus the GS of the given NHE is y = c_{1}e^{x} + c_{2}e^{–}^{x} – (1/2)(cos x + x sin x).
When f (x) Is A Product Of An Exponential Function And A Sine Or Cosine Function
Let's find a PS of the NHE:
y'' – 3y' + 2y = e^{x} sin x.
Let y = ae^{x}(A cos x + B sin x). Then y = e^{x}(aA cos x + aB sin x). So we can simply utilize y
= e^{x}(A cos x + B sin x). It
works ok, as evidenced in Example 4.10 below.
Example 4.10
Determine the GS of y'' – 3y' + 2y = e^{x} sin x.
Solution
EOS
Return To Top Of Page Go To Problems & Solutions
5. Modification Of The Method Of Undetermined Coefficients 
Let's find a PS of the NHE:
y'' + y' = x^{3} – x^{2}.
Let's try y = a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}. We have:
y = a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}, y' = 3a_{3}x^{2} + 2a_{2}x + a_{1}, y'' = 6a_{3}x + 2a_{2},
x^{3} – x^{2} = y''
+ y' = (6a_{3}x + 2a_{2}) + (3a_{3}x^{2} + 2a_{2}x + a_{1}) = 3a_{3}x^{2} + (6a_{3} + 2a_{2})x + 2a_{2} + a_{1},
3a_{3}x^{2} + (6a_{3} + 2a_{2})x + 2a_{2} + a_{1} = x^{3} – x^{2}.
We may add the term 0x^{3} to the
lefthand side, but then we can't equate 0 to the coefficient 1 of x^{3} on the righthand side, and
so the process of equating coefficients can't be done.
The trial solution y = a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} doesn't work. That's because when we compute (the xexpressions of ) y'
and y'',
they're polynomials of degree 2, and so is y''
+ y', while x^{3} – x^{2} is a
polynomial of degree 3. The term containing y, which is of
degree 3 in x, is absent from the lefthand side.
So what do we do? Ok, let's try a polynomial of degree 4, so
that when we compute y''
+ y', it'll be a
polynomial of degree 3,
the same as that of x^{3} – x^{2}. Let y = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}. We have:
y = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}, y' = 4a_{4}x^{3} + 3a_{3}x^{2} + 2a_{2}x + a_{1}, y'' = 12a_{4}x^{2} + 6a_{3}x + 2a_{2},
x^{3} – x^{2} = y''
+ y' = (12a_{4}x^{2} + 6a_{3}x + 2a_{2}) + (4a_{4}x^{3} + 3a_{3}x^{2} + 2a_{2}x + a_{1}) = 4a_{4}x^{3} + (12a_{4} + 3a_{3})x^{2} + (6a_{3} + 2a_{2})x + 2a_{2} + a_{1},
4a_{4}x^{3} + (12a_{4} + 3a_{3})x^{2} + (6a_{3} + 2a_{2})x + 2a_{2} + a_{1} = x^{3} – x^{2}.
The constant term a_{0} is absent from
the lefthand side, and thus can't be determined. The trial solution y = a_{4}x^{4} + a_{3}x^{3} + a_{2}x^{2}
+ a_{1}x + a_{0} doesn't work either.
So, again, what do we do? Ok, let's multiply the initial
trial solution a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0} by x to get the
modified trial solution
y = a_{3}x^{4} + a_{2}x^{3} + a_{1}x^{2} + a_{0}x. This trial PS works, as will be seen in
Example 5.1 below.
When f (x)
is a polynomial and y is absent
from the lefthand side (because its coefficient is 0), the modified trial
solution is a
polynomial of degree greater than that of the polynomial f (x)
by 1 and with the constant coefficient equal to 0. In the term
a_{i}x^{i}^{+1}, the
subscript i of the coefficient a_{i} is less than the
exponent i + 1 of x^{i}^{+1} by 1, to
reveal the multiplication by x and to
stop at a_{0}x where the
subscript is already the lowest 0. This is an example where the method of
undetermined coefficients
must be modified.
Find the GS of the DE y'' + y' = x^{3} – x^{2}.
Solution
a_{3} = 1/4, a_{2} = (– 1 – 12(1/4)) / 3 = –4/3, a_{1} = –6(–4/3)/2 = 4, a_{0} = –2(4) = –8,
a PS of the NHE is y = (1/4)x^{4} – (4/3)x^{3} + 4x^{2} – 8x.
So the GS of the given NHE is y
= c_{1} + c_{2}e^{–}^{x} + (1/4)x^{4} – (4/3)x^{3} + 4x^{2} – 8x.
EOS
Multiplying By x
Let's determine a PS of the NHE:
y'' – y = 2e^{x}.
Let's try y = ae^{x}. We have:
y = ae^{x}, y' = ae^{x}, y'' = ae^{x},
2e^{x}
= y'' – y = ae^{x}
– ae^{x} = 0.
The case of the polynomial in Example 5.1 suggests that we can try the multiplication of the
initial trial solution ae^{x}
by x. Let's
try y = axe^{x}.
This trial solution works quite well, as shown in Example 5.2 below. Note that axe^{x} isn't a PS of the HE
for any
values of c_{1} and c_{2}.
In the solution we of course first find the GS of the HE. If
we observe that the initial trial PS g(x) of the NHE is a PS of the HE,
then we multiply it by x, if xg(x) obtained
isn't a PS of the HE, then we use it as the modified trial PS. This is another
example
where the method of undetermined coefficients must be modified.
Solve y'' – y = 2e^{x}.
Solution
For a PS of the given NHE y''
– y = 2e^{x}.
The function ae^{x} is a PS of the HE and
the function axe^{x}
isn't. So let's try y = axe^{x}.
We have:
y = axe^{x}, y' = ae^{x} + axe^{x}, y'' = ae^{x} + (ae^{x} + axe^{x}) = 2ae^{x} + axe^{x},
2e^{x}
= y'' – y = (2ae^{x} + axe^{x})
– axe^{x} = 2ae^{x},
2ae^{x} = 2e^{x},
a = 1,
a PS is y = xe^{x}.
So the GS of the given NHE is y
= c_{1}e^{x} + c_{2}e^{–}^{x} + xe^{x},
or y = (x
+ c_{1})e^{x}
+ c_{2}e^{–}^{x}.
EOS
Solve the initialvalue problem y''
+ y = cos x, y(0) = 2, y'(0) = –3.
For a PS of the given NHE. Let's try y = A cos x + B sin x. We have:
y = A cos x + B sin x, y' = – A sin x + B cos x, y'' = – A cos x – B sin x,
cos x = y'' + y = (– A cos x – B sin x) + (A cos x + B sin x) = 0.
Solution
c_{1} = 2,
c_{2} = –3.
Thus the unique solution of the given initialvalue problem is y = 2 cos x – 3 sin x + (1/2)x sin x.
EOS
Let's find a PS of the equation:
y'' – 2y' + y = e^{x}.
Let's try y = ae^{x}. We have:
y = ae^{x}, y' = ae^{x}, y'' = ae^{x},
e^{x} = y'' – 2y' + y = ae^{x} – 2ae^{x} + ae^{x} = 0.
The trial PS y = ae^{x} fails. It's a PS of
the HE y'' – 2y' + y = 0, and thus can't be a PS of the NHE. Let's
determine the GS of the
HE. We have (–2)^{2}
– 4(1) = 0, –b/2 = –(–2)/2 = 1, and the GS is y = c_{1}e^{x} + c_{2}xe^{x}. The trial PS y = ae^{x}
of the NHE is a PS of
the HE, obtained from the GS of the HE by letting c_{1} = a and c_{2} = 0.
Now let's multiply the initial trial PS ae^{x}
by x to get axe^{x}.
But axe^{x} is
still a PS of the HE, obtained by letting c_{1} = 0 and c_{2} = a.
So y = axe^{x}
would fail too. Let's check:
y = axe^{x}, y' = ae^{x} + axe^{x}, y'' = ae^{x} + ae^{x} + axe^{x} = 2ae^{x} + axe^{x},
e^{x} = y'' – 2y' + y = 2ae^{x} + axe^{x} – 2(ae^{x} + axe^{x}) + axe^{x} = 0.
Indeed y = axe^{x} fails too.
So what do we do? Ok, let's multiply the initial trial PS ae^{x} by x^{2} to get ax^{2}e^{x}. Now ax^{2}e^{x}
isn't a PS of the HE. Thus let's try y =
ax^{2}e^{x}.
This trial solution works, as evidenced in Example 5.4 below. Note that ax^{2}e^{x}
isn't a PS of the HE for any values of c_{1}
and c_{2}. This is another example where the
method of undetermined coefficients must be modified.
In the solution we of course first find the GS of the HE. If
we observe that the initial trial PS g(x) of the NHE is a PS of the HE,
then we multiply it by x^{n},
where n is the smallest positive integer
such that x^{n}g(x) isn't a PS of the HE, and we use
the
modified trial PS y = x^{n}g(x).
Determine the GS of the DE y'' – 2y' + y = e^{x}.
Solution
For the GS of the HE y''
– 2y' + y = 0, we have (–2)^{2} – 4(1) = 0, –b/2 = –(–2)/2 = 1, and the GS is y = c_{1}e^{x} + c_{2}xe^{x}.
For a PS of the given NHE y''
– 2y' + y = e^{x}.
The functions y = ae^{x}
and axe^{x} are
PSs of the HE. So let's try y = ax^{2}e^{x}.
We
have:
y = ax^{2}e^{x}, y' = 2axe^{x} + ax^{2}e^{x}, y'' = (2ae^{x} + 2axe^{x}) + (2axe^{x} + ax^{2}e^{x}) = 2ae^{x} + 4axe^{x} + ax^{2}e^{x},
e^{x}
= y'' – 2y' + y = (2ae^{x}
+ 4axe^{x} + ax^{2}e^{x})
– 2(2axe^{x} + ax^{2}e^{x})
+ ax^{2}e^{x}
= 2ae^{x},
2ae^{x} = e^{x},
2a = 1,
a = 1/2,
a PS of the NHE is y = (1/2)x^{2}e^{x}.
Thus the GS of the given NHE is y = c_{1}e^{x} + c_{2}xe^{x} + (1/2)x^{2}e^{x}, or y = ((1/2)x^{2} + c_{2}x + c_{1})e^{x}.
When One Or More Terms Of The Initial Trial PS Of The NHE Is A PS Of The HE
Let's determine a PS of the equation:
y'' + y = x sin x.
Let's try y = (a_{1}x + a_{0}) cos x + (b_{1}x + b_{0}) sin x. We have:
y = (a_{1}x + a_{0}) cos x + (b_{1}x + b_{0}) sin x,
y' = a_{1} cos
x – (a_{1}x + a_{0}) sin
x + b_{1} sin
x + (b_{1}x + b_{0}) cos
x = (b_{1}x + a_{1} + b_{0}) cos x + (– a_{1}x – a_{0} + b_{1}) sin
x,
y'' = b_{1} cos x – (b_{1}x + a_{1} + b_{0}) sin x – a_{1} sin x + (– a_{1}x – a_{0} + b_{1}) cos x = (– a_{1}x – a_{0} + 2b_{1}) cos x + (– b_{1}x – 2a_{1} – b_{0}) sin x,
x sin
x = y''
+ y = 2b_{1} cos
x – 2a_{1} sin
x,
2b_{1} cos x – 2a_{1} sin
x = x sin
x.
Now the factor x in x sin x is a
polynomial of degree 1, but there's no polynomial of degree 1 as a factor in –
2a_{1} sin x. The
process of equating coefficients can't be done. The trial PS y = (a_{1}x + a_{0}) cos
x + (b_{1}x + b_{0}) sin
x doesn't work. Let's find the
GS of the HE y'' + y = 0. As done in Example 5.3, it's y = c_{1} cos
x + c_{2} sin
x. The trial PS is:
y = (a_{1}x + a_{0}) cos x + (b_{1}x + b_{0}) sin x = a_{1}x cos x + a_{0} cos x + b_{1}x sin x + b_{0} sin x.
Clearly the terms a_{0} cos
x and b_{0} sin
x are solutions of the HE. They disappear from
the xexpression of y''
+ y. So the trial
solution is the same as if it's y = a_{1}x cos
x + b_{1}x sin x. Let's
check:
y = a_{1}x cos x + b_{1}x sin
x,
y'
= a_{1} cos x – a_{1}x sin x + b_{1} sin x + b_{1}x cos
x = (b_{1}x + a_{1}) cos x + (– a_{1}x + b_{1}) sin x,
y''
= b_{1} cos x – (b_{1}x + a_{1}) sin
x – a_{1} sin x + (– a_{1}x + b_{1}) cos x = (– a_{1}x + 2b_{1}) cos x + (– b_{1}x – 2a_{1}) sin x,
y'' + y = 2b_{1} cos x – 2a_{1} sin x,
which is the same as found above. Indeed the trial solution
is the same as if it's y = a_{1}x cos
x + b_{1}x sin x, which is
incomplete, and consequently doesn't work.
Let's multiply the initial trial solution by x to get the modified trial solution (a_{1}x^{2} + a_{0}x) cos
x + (b_{1}x^{2} + b_{0}x) sin x.
No term of
this modified trial solution is a solution of the HE. This modified trial
solution works quite well, as shown in Example 5.5 below.
In the solution we of course first find the GS of the HE. If
we observe that one or more terms of the initial trial PS g(x) of the
NHE are PSs of the HE, then we multiply g(x) by x^{n},
where n is the smallest positive integer
such that no term of x^{n}g(x) is
a
solution of the HE, and we use the modified trial PS y
= x^{n}g(x).
Solve the equation y'' + y = x sin x.
Solution
b_{1} = 0, a_{1} = –1/4, b_{0} = –2(–1/4)/2 =
1/4, a_{0} = 0,
a PS of the NHE is y = (–1/4)x^{2} cos x + (1/4)x sin x.
So the GS of the given NHE is y
= c_{1} cos x + c_{2} sin
x + (–1/4)x^{2} cos
x + (1/4)x sin
x, or y = (c_{1} – (1/4)x^{2}) cos
x +
(c_{2} + (1/4)x)
sin x.
Let's recall the following terminology. In the expression:
y = ax^{2}e^{x} + bx sin 3x,
ax^{2}e^{x} and bx sin 3x are terms of y. In the expression:
y = ax^{2}e^{x},
a, x^{2} (or x, as x^{2} = xx), and e^{x} are factors of y.
In Example 5.5, the modified trial solution of the NHE is:
y = (a_{1}x^{2} + a_{0}x) cos x + (b_{1}x^{2} + b_{0}x) sin x = x^{2}(a_{1} cos x + b_{1} sin x) + x(a_{0} cos x + b_{0} sin x).
The items a_{1} cos x, b_{1} sin x, a_{0} cos x, and b_{0} sin x are solutions of the HE, but they cause no
problem, because they're not
terms of y. (The terms are a_{1}x^{2} cos x, b_{1}x^{2} sin
x, a_{0}x cos
x, and b_{0}x sin
x.)
In Example 5.2, the
initial trial solution is y = ae^{x}, which is a solution
of the HE. The modified trial solution is y = axe^{x} =
x(ae^{x}),
where the factor ae^{x} is
a solution of the HE, but it causes no problem, because it's not a term of y. Similarly for
Examples 5.3 and 5.4.
Return To Top Of Page Go To Problems & Solutions
6. Recap 
We now recap the results of this section. Consider the NHE:
y'' + by' + cy = f (x)
and its associated HE:
y'' + by' + cy = 0.
Case 1. No part of the trial PS of the NHE is a
solution of the HE. A PS of the NHE has the form of y
as described in the table
below, where P_{n}(x) and Q_{n}(x) are polynomials of degree n.
^{{6.1}} Examples 4.2 and 4.3
^{{6.2}} Example 5.1
^{{6.3}}
Example 4.4
^{{6.4}} Example
4.5
^{{6.5}}
Example 4.6
^{{6.6}}
Example 4.8
^{{6.7}}
Example 4.9
^{{6.8}}
Example 4.10
Case 2. One or more terms of the initial trial PS of
the NHE are solutions of the HE. Multiply the appropriate trial solution y in
the table above by x^{n},
where n is the smallest positive integer
such that no term of the obtained product is a solution of the HE.
Problems & Solutions 
1. Find the GS of y'' + 4y' + 4y = 12.
For the GS of the HE y'' + 4y' + 4y = 0, we have 4^{2} – 4(4) = 0, –b/2 = –4/2 = –2, and the GS is y = c_{1}e^{–2}^{x} + c_{2}xe^{–2}^{x}.
For a PS of the given NHE y''
+ 4y' + 4y = 12, let's try y
= 12/4 = 3. Then y' = 0, y'' = 0, and y''
+ 4y' + 4y = 0 + 4(0) + 4(3) =
12. So a PS of the NHE is y = 3.
Thus the GS of the NHE is y = c_{1}e^{–2}^{x} + c_{2}xe^{–2}^{x} + 3.
2. Determine the GS of y'' – 3y' + 2y = x + 1.
3. Find the unique solution of the initialvalue problem:
a_{2} = 1, a_{1} = 1, a_{0} = 1 – 2(1) =
–1,
a PS is y = 1x^{2} + 1x – 1 or y = x^{2} + x – 1.
So the GS of the NHE is y = c_{1} cos x + c_{2} sin x + x^{2} + x – 1. Then y' = – c_{1} sin x + c_{2} cos x + 2x + 1. We have:
3 = y(0) = c_{1} cos 0 + c_{2} sin 0 + 0^{2} + 0 – 1 = c_{1} – 1, c_{1} = 4,
1 = y'(0)
= – c_{1} sin 0 + c_{2} cos 0 + 2(0) + 1 = c_{2} + 1, c_{2} = 0.
Thus the required solution is y = 4 cos x + 0 sin x + x^{2} + x – 1, or y = 4 cos x + x^{2} + x – 1.
4. Solve y'' – 3y' + 2y = 6e^{3}^{x}.
Solution
For a PS of the given NHE y'' – 3y' + 2y = 6e^{3}^{x}, let y = ae^{3}^{x}. We have:
y = ae^{3}^{x}, y' = 3ae^{3}^{x}, y'' = 9ae^{3}^{x},
6e^{3}^{x}
= y'' – 3y' + 2y = 9ae^{3}^{x} – 3(3ae^{3}^{x}) + 2ae^{3}^{x} = 2ae^{3}^{x},
2ae^{3}^{x} = 6e^{3}^{x},
2a = 6,
a = 3,
a PS is y = 3e^{3}^{x}.
So the required GS is y = c_{1}e^{2}^{x} + c_{2}e^{x} + 3e^{3}^{x}.
5. Find the GS of the equation y'' + 4y = 3 sin x.
A
= 0, B = 1,
a PS is y = 0 cos x + 1 sin x
or y = sin x.
So the GS of the given NHE is y = c_{1} cos 2x + c_{2} sin 2x + sin x.
6. Determine the GS of the DE y''
– 2y' = 2 sin
x + 2e^{3}^{x}.
Hint: use the principle of superposition.
Solution
For a PS of the NHE y'' – 2y' = 2 sin x, let y = A cos x + B sin x. We have:
y = A cos x + B sin x, y' = – A sin x + B cos x, y'' = – A cos x – B sin x,
2 sin x = y'' – 2y'
= (– A cos x – B
sin x) – 2(– A
sin x + B
cos x) = (2A
– B) sin x + (– A
– 2B) cos x,
(2A – B) sin x
+ (– A – 2B) cos x
= 2 sin x,
_{
}
7. Solve the initialvalue problem y'' – 4y' + 4y = 6xe^{3}^{x}, y(0) = 0, y'(0) = 3.
Solution
For the GS of the HE y'' – 4y' + 4y = 0, we have (–4)^{2} – 4(4) = 0, –b/2 = –(–4)/2 = 2, and the GS is y = c_{1} e^{2}^{x} + c_{2}xe^{2}^{x}.
For a PS of the given NHE, let y = (a_{1}x + a_{0})e^{3}^{x}. We have:
y = (a_{1}x + a_{0})e^{3}^{x},
y'
= a_{1}e^{3}^{x}
+ 3(a_{1}x + a_{0})e^{3}^{x}
= (a_{1} + 3a_{0})e^{3}^{x} + 3a_{1}xe^{3}^{x},
y''
= 3(a_{1} + 3a_{0})e^{3}^{x} + 3a_{1}e^{3}^{x}
+ 9a_{1}xe^{3}^{x}
= (6a_{1} + 9a_{0})e^{3}^{x} + 9a_{1}xe^{3}^{x},
6xe^{3}^{x} = y'' – 4y'
+ 4y = (6a_{1} + 9a_{0})e^{3}^{x}
+ 9a_{1}xe^{3}^{x}
– 4((a_{1} + 3a_{0})e^{3}^{x} + 3a_{1}xe^{3}^{x})
+ 4(a_{1}x + a_{0})e^{3}^{x}
= (a_{1}x + 2a_{1} + a_{0})e^{3}^{x},
(a_{1}x + 2a_{1} + a_{0})e^{3}^{x} = 6xe^{3}^{x},
a_{1}x + 2a_{1} + a_{0} = 6x,
a_{1} = 6, a_{0} = –2(6) = –12,
a PS of the NHE is y = (6x – 12)e^{3}^{x},
or y = 6(x
– 2)e^{3}^{x}.
So the GS of the NHE is y = c_{1}e^{2}^{x} + c_{2}xe^{2}^{x} + 6(x – 2)e^{3}^{x}. We have:
0 = y(0) = c_{1}e^{0} + 0 + 6(0 – 2)e^{0} = c_{1} – 12, c_{1} = 12,
y'
= 2c_{1}e^{2}^{x}
+ c_{2}e^{2}^{x}
+ 2c_{2}xe^{2}^{x}
+ 6e^{3}^{x} + 18(x – 2)e^{3}^{x}
= (2c_{1} + c_{2})e^{2}^{x} + 2c_{2}xe^{2}^{x}
+ (18x – 30)e^{3}^{x},
3 = y'(0) = (2(12) + c_{2})e^{0} + 0 + (0 – 30)e^{0} = c_{2} – 6, c_{2} = 9.
Thus the unique solution of the given initialvalue problem is y = 12e^{2}^{x} + 9xe^{2}^{x} + 6(x – 2)e^{3}^{x}.
8. Solve the NHE y'' + 4y = 16x sin 3x.
Solution
9. Find the GS of the NHDE y'' + 4y' + 4y = e^{2}^{x} cos x.
Solution
For the GS of the HE y'' + 4y' + 4y = 0, we have 4^{2} – 4(4) = 0, –b/2 = –4/2 = –2, and the GS is y = c_{1}e^{–2}^{x} + c_{2}xe^{–2}^{x}.
For a PS of the given NHE, let y = e^{2}^{x}(A cos x + B sin x). We have:
y = e^{2}^{x}(A cos
x + B
sin x),
y'
= 2e^{2}^{x}(A cos
x + B
sin x) + e^{2}^{x}(–
A sin x + B cos x),
y''
= 4e^{2}^{x}(A cos
x + B
sin x) + 2e^{2}^{x}(–
A sin x + B cos x) + 2e^{2}^{x}(– A sin
x + B
cos x) + e^{2}^{x}(–
A cos x – B sin x)
= 4e^{2}^{x}(A cos
x + B
sin x) + 4e^{2}^{x}(–
A sin x + B cos x) – e^{2}^{x}(A cos
x + B
sin x)
= 3e^{2}^{x}(A cos
x + B
sin x) + 4e^{2}^{x}(–
A sin x + B cos x),
4y' = 8e^{2}^{x}(A cos
x + B
sin x) + 4e^{2}^{x}(–
A sin x + B cos x),
4y = 4e^{2}^{x}(A cos x + B sin x),
e^{2}^{x} cos x = y'' + 4y'
+ 4y
= 15e^{2}^{x}(A cos x + B sin x) + 8e^{2}^{x}(– A sin
x + B
cos x)
= e^{2}^{x}((15A + 8B) cos x + (– 8A + 15B) sin x),
e^{2}^{x}((15A + 8B) cos x + (– 8A + 15B) sin x) = e^{2}^{x} cos x,
(15A + 8B) cos x
+ (– 8A + 15B) sin x
= cos x,
[1] x 8: 120A + 64B
= 8, [3]
[2] x 15: – 120A + 225B
= 0, [4]
[3] + [4]: 289B = 8, B
= 8/289,
[1]: A = (1 – 8(8/289))/15 = 15/289,
a PS of the NHE is y = e^{2}^{x}((15/289) cos x + (8/289) sin x).
So the GS of the given NHE is y = c_{1}e^{–2}^{x} + c_{2}xe^{–2}^{x} + e^{2}^{x}((15/289) cos x + (8/289) sin x).
10. Determine the GS of the equation y'' + y' = 3x^{2}.
a_{2} = 3/3 = 1, a_{1} = –6(1)/2 = –3, a_{0} = –2(–3) = 6,
a PS of the NHE is y = x^{3} – 3x^{2} + 6x.
So the GS of the given NHE is y = c_{1} + c_{2}e^{–}^{x} + x^{3} – 3x^{2} + 6x.
11. Solve the DE y'' + 4y = sin 2x.
B
= 0, A = –1/4,
a PS of the NHE is y = –(1/4)x cos 2x.
Thus the GS of the given NHE is y = c_{1} cos 2x + c_{2} sin 2x – (1/4)x cos 2x, or y = (– (1/4)x + c_{1}) cos 2x + c_{2} sin 2x.
12. Find the GS of x'' – 4x' + 4x = e^{2}^{t}.
For the GS of the HE x'' – 4x' + 4x = 0, we have (–4)^{2} – 4(4) = 0, –(–4)/2 = 2, and the GS of the HE is x = c_{1}e^{2}^{t} + c_{2}te^{2}^{t}.
For a PS of the given NHE. The functions x = ae^{2}^{t} and x = ate^{2}^{t} are PSs of the HE. So let's try x = at^{2}e^{2}^{t}. We have:
x = at^{2}e^{2}^{t},
x'
= 2ate^{2}^{t} +2at^{2}e^{2}^{t} = (2at^{2} + 2at)e^{2}^{t},
x''
= (4at + 2a)e^{2}^{t} + 2(2at^{2} + 2at)e^{2}^{t} = (4at^{2} + 8at + 2a)e^{2}^{t},
–4x' = (– 8at^{2} – 8at)e^{2}^{t},
4x = 4at^{2}e^{2}^{t},
e^{2}^{t} = x'' – 4x'
+ 4x = 2ae^{2}^{t},
2ae^{2}^{t} = e^{2}^{t},
2a = 1,
a = 1/2,
a PS of the NHE is x = (1/2)t^{2}e^{2}^{t}.
Thus the GS of the given NHE is x = c_{1}e^{2}^{t} + c_{2}te^{2}^{t} + (1/2)t^{2}e^{2}^{t}, or x = ((1/2)t^{2} + c_{2}t + c_{1})e^{2}^{t}.
y'' + by' + cy = f (x)
has a PS that's a polynomial of degree n.
Solution
Suppose:
14. Prove that the GS of the NHDE:
Solution
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