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1. SecondOrder Linear Homogeneous Differential Equations With 
In this section, we'll use the abbreviations:
DE 
for 
differential equation, 
GS 
for 
general solution, 
HDE 
for 
homogeneous differential equation, and 
HE 
for 
homogeneous equation. 
Recall that a secondorder linear homogeneous differential equation with constant coefficients is one of the form:
y'' + by' + cy = 0,
where the coefficients b and c are
constants. When b is replaced by a
nonconstant function b(x) or c is replaced
by a
nonconstant function c(x) or both, one or both coefficients of course
vary, they become variable. So naturally an equation of
the form:
y'' + b(x) y' + c(x) y = 0,
where b(x) or c(x) or both are nonconstant functions of x, is said to be an equation with variable
coefficients. Note that x is
the independent variable of the function y, and thus
of the functions y' and y'' also. All the functions in the equation
are of the
same variable.
Definition 1.1 SecondOrder Linear Homogeneous
Differential Equations With Variable
Coefficients
A differential equation of the form: where b(x) or c(x) or both are nonconstant functions, is
called a secondorder linear homogeneous 
Recall that while the equation is linear, each function y, y',
and y'' 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 Solution 
Just like the GS of the constantcoefficients HE:
y'' + by' + cy = 0
as presented in Section 16.2.1 Theorem 2.2, the GS of the variablecoefficients HE:
y'' + b(x) y' + c(x) y = 0
is the linear combination of any two of its linearly
independent PSs. The proof of the following theorem is given in a course on
differential equations and is omitted here.
Theorem 2.1 General
Solution
If y_{1} and y_{2} are linearly independent particular solutions of the HE: y'' + b(x) y' + c(x) y = 0, then the GS of this equation is: y = c_{1} y_{1} + c_{2} y_{2}, where c_{1} and c_{2} are arbitrary constants. 
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3. The Method Of Reduction Of Order 
Recall that to solve a DE is to find its GS. Let's discuss the solving of our equation:
By Theorem 2.1, we obtain the GS
of Eq. [3.1] provided we get two linearly independent solutions y_{1} and y_{2} of this
equation.
The GS is then given by the linear combination:
y = c_{1} y_{1} + c_{2} y_{2},
where c_{1} and c_{2} are arbitrary constants. Clearly y_{1} = 0 is a solution.
But it's not linearly independent of any other solution (0 =
0y_{2} whatever y_{2} is). So of
course we're interested in non0 solutions.
Unlike the case of constantcoefficients HEs as developed in
Section
16.2.1 Part 3, there's no standard procedure to find two
linearly independent solutions y_{1} and y_{2} of the variablecoefficients HE [3.1]. However, there's
a general procedure for finding
another solution if one solution is known. It's often possible to find one
solution by inspecting the equation or by trial and error.
Let y_{1} be a non0
solution of Eq. [3.1]. If y_{2} is another
solution that's linearly independent of y_{1}, then y_{2}/y_{1} isn't a
constant.
Thus:
Indeed y_{2} is a solution. Now let's verify that it's linearly independent of y_{1}. We have:
We've reduced a secondorder equation to a firstorder
equation. The order of the derivative is reduced. So this approach of
using one solution to find another is called the method of reduction of
order.
We've of course proved the following theorem.
Theorem 3.1 A
Solution By The Method Of Reduction Of Order
Suppose y_{1} is a solution of the DE: where c_{1} and c_{2} are arbitrary constants. 
Consider the DE:
x^{2}y'' xy' + y = 0, x > 0,
a. Verify that y_{1} = x is a solution.
b. Find a second solution y_{2} that's
linearly independent of y_{1} by using the
formula obtained by the method of reduction of order.
c. Find y_{2} by imitating the method of reduction
of order itself, not using the formula obtained by it.
d. Are the answers in parts b and c the same?
e. Verify that y_{2} is indeed a
solution that's linearly independent of y_{1}.
Solution
xu'' + u'
= 0,
divide both sides by xu':
y_{2} = uy_{1} = (ln x) x = x ln x.
d. Yes.
Since ln x isn't a constant, y_{2} is linearly independent of y_{1}. Thus indeed y_{2} is a solution that's linearly independent of y_{1}.
a. In part b, we utilize the formula for y_{2}, which contains the function b(x), so we should write the given equation in the
form
y'' + b(x) y' + c(x) y
= 0 (coefficient of y'' is 1), so that
it's clear what b(x)
is. Caution: b(x) isn't x, it's 1/x.
b. Part c illustrates the following general procedure of the method of reduction of order:
let y_{2} = uy_{1} = g(x)u, where y_{1} = g(x), use g(x) only, not y_{1},
substitute y_{2} = g(x)u into the original equation,
transform the term containing u'' in an equation into the term u''/u'
by division,
let v = u',
find v,
then u, then y_{2}, see Fig. 3.1.

Fig. 3.1 Process From y_{2} To u To v Then Back To u To y_{2}. 
d. If we're not asked to verify a solution, then we don't have to verify.
It's recommended that you remember the procedure of the
method of reduction of order, as outlined in part b of Remarks 3.2
above. In case you forget its formula or don't remember the formula clearly,
you can utilize the procedure of the method.
Sometimes you may even be required to use the method itself, not the formula
obtained by it.
Choosing y_{2}
Suppose it's determined that y_{2} is a constant
multiple of a function, ie that it's of the form y_{2} = k f (x), where k is a non0
constant. For example, y_{2} = 5x^{3} or y_{2} = 3(sin
2x + 7e^{4}^{x}).
We now verify that f (x)
is also a solution that's linearly independent
of y_{1}. We have:
which isn't constant, since y_{2}/y_{1} isn't. Thus f (x) is linearly independent of y_{1}. That completes the verification.
We've shown that if k f (x),
where k is a non0 constant, is a solution
that's linearly independent of y_{1}, then so is f (x). We
conclude that if we determine that y_{2} = k f (x), then it's
possible to simply choose y_{2} = f (x).
Let's look at this possibility from another angle. If y_{2} = k f (x),
then the GS is y = c_{1}y_{1} + c_{2}k f (x). Reusing
the letter c_{2} for the
product c_{2}k, which is
valid because this product, just like the factor c_{2} in it, is an
arbitrary constant, the GS is y = c_{1}y_{1} +
c_{2} f (x).
This is the same as simply choosing y_{2} = f (x).
One solution of the DE:
xy'' y' + 4x^{3}y = 0, x > 0,
is y_{1} = sin (x^{2}). Solve the equation.
Solution
xy'' y'
+ 4x^{3}y = 0,
Let u = x^{2}. So du = 2x dx. Then:
So, choosing y_{2} = cos
(x^{2}), the GS is y
= c_{1} sin (x^{2}) + c_{2} cos
(x^{2}).
EOS
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4. For ConstantCoefficients Equations 
The procedure of the method of reduction of order doesn't require
that the coefficients be variable functions. So the method
also applies to constantcoefficients equations.
Consider the constantcoefficients DE:
y'' + 5y' + 6y = 0.
a. Solve it by using the method of characteristic
equation.
b. Given that one solution is y_{1} = e^{2}^{x}, solve it by using the
method of reduction of order.
c. Are the answers in parts a and b the same?
Solution
So, choosing y_{2} = e^{3}^{x}, the GS is y = c_{1}e^{2}^{x} + c_{2}e^{3}^{x}.
c. Yes.
EOS
Problems & Solutions 
1. Consider the Legendre equation of order 1:
(1 x^{2}) y'' 2xy' + 2y = 0, x < 1.
a. Verify that y_{1} = x is a solution.
b. Find a
second solution y_{2} that's linearly independent of y_{1} by using the formula obtained by the method of reduction
of order.
c. Find y_{2} by imitating
the method of reduction of order itself, not using the formula obtained by it.
d. Are the answers in parts b
and c the same?
e. Verify that y_{2} is indeed a solution that's linearly independent of y_{1}.
Let u = 1 x^{2}. So du = 2x dx. Then:
Using the method of partial fractions we have:
C = 0,
D
= 1,
0 = A + B
+ C = A + B + 0 = A
+ B, B
= A,
0 = A + B
+ D = A A 1 = 2A
1, A = 1/2,
B
= (1/2) = 1/2,
Using the method of partial fractions we have:
Using the method of partial fractions we have:
d. Yes.
e) Let's show that y_{2} is a solution:
So indeed y_{2} is a solution. Now let's show that it's linearly independent of y_{1}:
which clearly isn't constant. Thus indeed y_{2} is linearly independent of y_{1}.
2. Consider the DE:
x^{2}y'' 2xy' + (x^{2} + 2) y = 0, x > 0.
a. Verify
that y_{1} = x sin
x is a solution.
b. Solve the equation.
Choose y_{2} = x cos x. So the GS of the given equation is y = c_{1} x sin x + c_{2} x cos x, or:
y = x(A cos
x + B sin
x),
where A = c_{2} and B = c_{1} are arbitrary constants.
3. Consider the constantcoefficients DE:
y'' + y' + 7y = 0.
a. Verify that:
is a solution.
b. Use y_{1} to solve the equation.
We're required to use y_{1} to solve the equation.
So we can't apply the method of characteristic equation, which doesn't use y_{1}.
We must apply the method of reduction of order, which uses y_{1} to find y_{2}.
Indeed y_{1} is a solution.
b. Let y_{2} be another solution that's linearly independent of y_{1}. Then by the method of reduction of order we have:
where A = c_{2} and B = c_{1} are arbitrary constants.
4. Consider the Bessel DE:
x^{2}y'' + xy' + (x^{2} p^{2}) y = 0, p a constant.
For p = 1/2 and x > 0:
b. Solve the equation.
a. When p = 1/2 the given equation becomes:
where A and B are arbitrary constants.
5. Given the DE:
xy'' (x + n) y' + ny = 0, x > 0, integer constant n > 0,
verify that y_{1} = e^{x} is a solution and determine another solution linearly independent of y_{1}.
Solution
Let's verify that y_{1} is a solution. We have:
Thus:
I_{n} = x^{n}e^{}^{x} + nI_{n}_{1}
= e^{}^{x}(x^{n}) + n( x^{n}^{1}e^{}^{x} + (n 1)I_{n}_{2})
= e^{}^{x}(x^{n} + nx^{n}^{1}) + n(n 1)( x^{n}^{2}e^{}^{x} + (n 2)I_{n}_{3})
= e^{}^{x}(x^{n} + nx^{n}^{1} + n(n 1)x^{n}^{2}) + n(n 1)(n 2)( x^{n}^{3}e^{}^{x} + (n 3)I_{n}_{4})
= e^{}^{x}(x^{n} + nx^{n}^{1} + n(n 1)x^{n}^{2} + n(n 1)(n 2)x^{n}^{3}) + n(n 1)(n 2)(n 3)( x^{n}^{4}e^{}^{x} + (n 4)I_{n}_{5})
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