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1. Second-Order Linear Homogeneous Differential Equations With |
In this section, we'll use the abbreviations:
“CE” |
for |
“characteristic equation”, |
“DE” |
for |
“differential equation”, |
“GS” |
for |
“general solution”, and |
“PS” |
for |
“particular solution”. |
In the differential equation:
y'' + b(x)y' + c(x)y = q(x),
For example, the equations:
y'' + 3y' – 2y = 0 and y'' – 5x^{2}y' + 3e^{5}^{x}y = 0
are linear, while the equations:
y'' + 3( y')^{3} – 2y^{2} = 0 and y''y – 5x^{2}y' + 3e^{5}^{x}y = 0
are non-linear.
The differential equation y''
= f (x), as
discussed in Section
16.1.1 Part 3, is a second-order linear one with constant
coefficients. For this equation, b(x) = c(x) = 0 and q(x) = f (x).
Definition 1.1 – Second-Order Linear Homogeneous
Differential Equations With Constant
Coefficients
A differential equation of the form: where b and c are constants, is called a second-order linear homogeneous differential equation with constant coefficients. |
The DE:
ay'' + b_{1}y' + c_{1}y = 0,
where a isn't 1 or 0, can be transformed into one of the form of Eq. [1.1] by dividing both sides by a:
So y_{1} is also a
solution of ay'' + b_{1} y' + c_{1} y
= 0. Thus to solve ay''
+ b_{1} y'
+ c_{1} y = 0, where a isn't 1 or 0, we first divide both
sides of it by a to get the form of Eq.
[1.1].
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2. General Solution |
Of course the second-order linear homogeneous DE with constant coefficients:
always has y = 0 as a PS
(if y = 0 then y'
= 0 and y'' = 0 and
thus y'' + by' + cy = 0). Of
course we're interested in finding
non-0 solutions of this DE.
By Section
16.1.3 A Particular Case, the GS of the first-order linear DE y' + by = 0 is y = Ce^{–}^{bx},
where C is an arbitrary
constant. So a PS is y = e^{–}^{bx}. Thus it's natural to
wonder if Eq. [2.1] has a PS of the form:
The quadratic equation:
Definition 2.1 – The Characteristic Equation
Given the differential equation: is called the characteristic equation of the differential equation [2.2]. |
So the sum c_{1}y_{1} + c_{2}y_{2} is also a solution. We've proved the following theorem.
Suppose y_{1} and y_{2} are solutions of: Then for any constants c_{1} and c_{2}, the function: y_{3} = c_{1} y_{1} + c_{2} y_{2} is also a solution of Eq. [2.4]. |
Some PSs of the DE y'
= 2x are y
= x^{2}, y = x^{2} + 10, or y = x^{2} – 5/4. Its GS is y = x^{2} + C,
where C is an
arbitrary constant.
Recall that the GS of a DE is the solution that represents all the solutions,
ie, every PS can be obtained from the GS by giving
appropriate value(s) to the constant(s) present in the GS. This definition of
course also applies to the GS of a second-order
linear homogeneous DE with constant coefficients.
Linear Independence And General Solution
Let y_{1} and y_{2} be 2 different solutions of Eq. [2.4]. Then, by Theorem
2.1, for any constants c_{1} and c_{2} the function c_{1}y_{1} + c_{2}y_{2} is
also a solution. That is, if a function is of the form c_{1}y_{1} + c_{2}y_{2} then it's a solution. We wonder if the converse is true:
if a
function is a solution, is it of the form c_{1}y_{1} + c_{2}y_{2}? That is, is every solution of the form c_{1}y_{1} + c_{2}y_{2}? That means,
is c_{1}y_{1} + c_{2}y_{2}
the GS? That is, can every solution be obtained from c_{1} y_{1} + c_{2} y_{2} by giving
appropriate particular values to c_{1} and c_{2}?
Suppose y_{1}, y_{2}, and y_{3} are solutions
where y_{2} is a constant multiple of y_{1} but y_{3} isn't a
constant multiple of y_{1}. So y_{2} = ky_{1},
where k is a constant other than 0 and 1.
We have:
c_{1}y_{1} + c_{2}y_{2} = c_{1}y_{1} + c_{2}(ky_{1}) = (c_{1} + c_{2}k)y_{1} = Ky_{1},
where K
= c_{1} + c_{2}k, and any solution obtained from y_{1} and y_{2} can actually
be obtained from y_{1} alone. Now c_{1}y_{1} + c_{3}y_{3} can't be
reduced to the form Ly_{1}, where L
is constant, and any solution obtained from y_{1} and y_{3} must indeed be obtained from both y_{1}
and y_{3}; it can't be obtained just from y_{1} alone.
Thus a PS that's obtained from y_{1} and y_{3} where y_{3} isn't a
constant multiple of y_{1} can't be
obtained from y_{1} alone, ie it can't be
obtained from y_{1} and y_{2} where y_{2} is a constant multiple of y_{1}. Consequently c_{1}y_{1} + c_{2}y_{2} isn't the GS
if y_{2} is a constant multiple
of y_{1}.
Also recall that if two vectors v_{1} and v_{2} in a plane are linearly independent, then any vector v_{3} in the same plane can be
expressed as a linear combination of v_{1} and v_{2}: v_{3} = k_{1}v_{1} + k_{2}v_{2} for some
constants k_{1} and k_{2}. That is, the
linear combination
c_{1}v_{1} + c_{2}v_{2}, where c_{1} and c_{2} are arbitrary
constants, represents all the vectors in that plane.
Analogously, two functions f (x)
and g(x) are said
to be linealy independent if none is a constant multiple of the other, ie if
there doesn't exist a constant k such that f (x) = kg(x) for all x where both f (x) and g(x) are defined, ie if the ratio f (x)/g(x)
isn't a constant, or if they have the following property: if c_{1} f (x)
+ c_{2} g(x) = 0, where c_{1} and c_{2} are constants, then c_{1} = c_{2} = 0.
The answer to our wondering is as follows. If y_{1} and y_{2} are particular
linearly independent solutions of Eq. [2.4], then the GS
of Eq. [2.4] is y = c_{1}y_{1} + c_{2}y_{2}, where c_{1} and c_{2} are arbitrary constants. The GS is a linear combination
of the 2 particular
linearly independent solutions. The proof of the following theorem is given in
a course on differential equations and is omitted
here.
Theorem 2.2 – General
Solution
If y_{1} and y_{2} are linearly independent particular solutions of the HE: then the general solution of this equation is: y = c_{1} y_{1} + c_{2} y_{2}, where c_{1} and c_{2} are arbitrary constants. |
How about y = c_{1} y_{1} + c_{2} y_{2} + C, where C is also an arbitrary constant? Shouldn't it
instead be the general solution? Let's find
out:
So y = c_{1} y_{1} + c_{2} y_{2} + C isn't the general solution
of y'' + by' + cy = 0. The
reason is that the general solution of a
second-order equation involves only two arbitrary constants, not three, and y = c_{1} y_{1} + c_{2} y_{2} already involves two, which are c_{1}
and c_{2} .
Combining Theorems 2.1 and 2.2 we see that if y_{1} and y_{2} are solutions
of y'' + by' + cy = 0, then
for any constants c_{1} and c_{2},
c_{1} y_{1} + c_{2} y_{2} is also a
solution, and additionally if y_{1} and y_{2} are linearly independent, then for arbitrary constants c_{1} and c_{2} , c_{1} y_{1}
+ c_{2} y_{2} is the GS.
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3. The Method Of Characteristic Equation |
Recall that to solve a DE is to find its GS. Let's discuss the solving of our equation:
If b^{2} – 4c > 0, then the CE of Eq. [3.1] has 2 distinct real roots:
where c_{1} and c_{2} are arbitrary constants.
If b^{2} – 4c = 0, then the CE of Eq. [3.1] has 2 equal real roots:
divide both sides by u'y_{1}:
y_{2} = uy_{1} = (Cx + K)e^{–(}^{b}^{/2)}^{x}.
Let's check that y_{2} is indeed a solution of Eq. [3.1]:
y = c_{1}e^{–(}^{b}^{/2)}^{x} + c_{2}xe^{–(}^{b}^{/2)}^{x},
where c_{1} and c_{2} are arbitrary constants.
If b^{2} – 4c < 0, then the CE of Eq. [3.1] has 2 imaginary roots:
where c_{1} and c_{2} are arbitrary constants.
In Section 15.3 Problem & Solution 8, we established Euler formula:
e^{ix} = cos x + i sin x.
Thus the GS can be written as:
where c_{1} and c_{2} are arbitray constants.
All the above discussion shows that we've proved the following
theorem. The above procedure relies on the characteristic
equation, and thus is called the method of characteristic equation.
Recall that the GS of a DE is such that every PS of the
DE can be obtained from it by assigning appropriate values to the constants
present in it.
Theorem 3.1 – General
Solution Of y'' + by' + cy = 0
Consider the second-order linear homogeneous differential equation with constant coefficients: and let c_{1} and c_{2} be arbitrary constants. By the method of characteristic equation: and the general solution of the differential equation is: |
a. Once we've found 2 linearly independent solutions of Eq. [3.2], we've essentially found all the solutions of that equation.
b. The GS of Eq. [3.2] always involves the
exponential function e^{kx},
where k is a real number.
c. Eq. [3.2] has infinitely many solutions.
Find the GS of the DE y'' – 3y' + 2y = 0.
Solution
(–3)^{2} – 4(2) = 1,
the GS is:
y = c_{1}e^{2}^{x} + c_{2}e^{x},
where c_{1} and c_{2} are arbitrary constants.
EOS
Recall that to solve a DE is to find its GS.
Solution
where c_{1} and c_{2} are arbitrary constants.
EOS
When the letter x is used for
the function, of course we can't use it also for the variable, we must use
another letter. Usually
when the function is x, it's
customary that the variable is t.
Solve the equation w'' + w' + w = 0.
Solution
1^{2} – 4(1) = –3,
where c_{1} and c_{2} are arbitrary constants.
EOS
When a letter other than x, whatever it is, is used for the function, we can always use the letter x for the variable.
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4. For Initial-Value Problems |
As seen in Section
16.1.1 Remarks 4.1, an initial-value problem involving a first-order DE
specifies 1 initial condition. As seen in
the same
section Remarks 4.2, an initial-value problem involving a second-order DE
specifies 2 initial conditions. The DE y''
+
by' + cy is of
second order, and so an intial-value problem involving it specifies 2 initial
conditions. This fact is also clear from
the form of the equation's GS: y = c_{1}y_{1} + c_{2}y_{2}, where there
are 2 constants c_{1} and c_{2} to be determined
to get a PS. To solve an
initial-value problem is to find the PS of the DE that satisfies the
conditions.
Solve the initial-value problem:
Solution
3^{2} – 4(–10) = 49,
EOS
In Example 3.1, we differentiate the GS, use the condition y'(0) = 3 to get an equation in c_{1} and c_{2}, use the
condition y(0) = 1
to get another equation in c_{1} and c_{2}, obtain a system of 2 simultaneous equations in 2
unknowns c_{1} and c_{2}, solve it,
obtain a
unique set of solutions, and thus obtain a unique PS of the given initial-value
problem.
In general, the second-order linear DE y''
+ by' + cy = 0 has infinitely many PSs, but when 2
initial conditions are imposed,
there's only 1 PS that satisfies the conditions. Note that for a second-order
equation, there are 2 initial conditions, which are on
y' and y (= y^{(0)}).
The solution of the initial-value problem: where x_{0}, k_{0}, and k_{1} are real numbers, exists and is unique. |
This theorem is a special case where n
= 2 of the general theorem that asserts that the solution of an initial-value
problem
involving an nth-order linear DE:
y^{(}^{n}^{)} + p_{n}_{–1}(x)y^{(}^{n}^{–1)} + p_{n}_{–2}(x)y^{(}^{n}^{–2)} + ... + p_{2}(x)y'' + p_{1}(x)y' + p_{0}(x)y = q(x)
and n initial conditions:
y(x_{0}) = k_{0}, y'(x_{0}) = k_{1}, y''(x_{0}) = k_{2}, ..., y^{(}^{n}^{–2)}(x_{0}) = k_{n}_{–2}, y^{(}^{n}^{–1)}(x_{0}) = k_{n}_{–1},
exists and is unique provided the functions p_{i}(x)'s
and q(x) are
continuous on a neighborhood of x_{0}. This general
theorem is
presented and proved in a course on differential equations. Remark that for an nth-order equation, there are n conditions,
which are on y (= y^{(0)}) thru to y^{(}^{n}^{–1)}.
Solve the following initial-value problem:
Solution
0^{2} – 4(1) = – 4,
c_{1} = –1,
the desired PS is y = (–1) cos
x + (0) sin x,
or y = – cos x.
EOS
Problems & Solutions |
1. Find the GS of the DE 4y'' + 12y' + 5y = 0.
y = c_{1}e^{–(1/2)}^{x} + c_{2}e^{–(5/2)}^{x}.
2. Solve the DE x'' – 8x' + 16x = 0.
(–8)^{2} – 4(16) = 0,
3. Solve the equation y'' + 3y = 0.
0^{2} – 4(3) = – 12,
4. Find the solution of the initial-value problem:
(–5)^{2} – 4(6) = 1,
y = c_{1}e^{3}^{x}
+ c_{2}e^{2}^{x},
1 = y(0) = c_{1}e^{0} + c_{2}e^{0} = c_{1} + c_{2}, c_{1} + c_{2} = 1, [1]
y' = 3c_{1}e^{3}^{x} + 2c_{2}e^{2}^{x},
–1 = y'(0) = 3c_{1}e^{0} + 2c_{2}e^{0} = 3c_{1} + 2c_{2}, 3c_{1} + 2c_{2} = –1, [2]
[1] x (–3): – 3c_{1} – 3c_{2} = –3, [3]
[2] + [3]: – c_{2} = – 4,
c_{2} = 4,
[1]: c_{1} = 1 – c_{2} = 1 – 4 = –3,
y = – 3e^{3x} + 4e^{2x}, or y = 4e^{2x} – 3e^{3x}.
5. Solve the following initial-value problem:
Solution
4v'' + 20v' + 25v = 0,
6. Solve this initial-value problem:
Solution
0^{2} – 4(4) = –16,
c_{2} = –4/2 = –2,
y = 3 cos 2x
– 2 sin 2x.
7. Consider the DE:
y' + y^{2} + cy + d = 0,
where c and d are constants.
It's a first-order non-linear equation. Observe that the exponents of y decrease from 2 to 1 to
0 and that the coefficient of y^{2} is 1.
a. Transform the given y-equation
into a second-order linear homogeneous z-equation with
constant coefficients by letting z
be a function such that z'/z = y.
b. Prove that if z is the GS of the z-equation found in part a and y = z'/z, then y is the GS of the y-equation.
c. Prove that the GS of the y-equation, which is of first-order, indeed involves only 1 arbitrary constant.
Solution
a. Let z be a function such that z'/z = y. Then:
Thus, by part a, w is a solution of the z-equation,
and, by part b, w is a PS of the z-equation, for if it's the GS of the
z-equation
then v would be the GS of the y-equation, a contradiction. Since z is the GS of the z-equation
which is
second-order linear homogeneous with
constant coefficients, we have:
z = c_{1}z_{1} + c_{2}z_{2},
where z_{1} and z_{2} are linearly independent PSs of the z-equation and c_{1} and c_{2} are arbitrary constants. As a consequence:
w = k_{1}z_{1} + k_{2}z_{2},
where k_{1} and k_{2} are particular constants. Then:
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