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1. Second-Order Linear Non-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”, |
“HE” |
for |
“homogeneous equation”, |
"NHDE" |
for |
"non-homogeneous differential equation", |
"NHE" |
for |
"non-homogeneous equation", and |
"PS" |
for |
"particular solution". |
Recall that a second-order linear homogeneous differential equation with variable coefficients is one of the form:
y'' + b(x) y' + c(x) y = 0,
where b(x) or c(x) or both are non-constant functions of x. So of course when the right-hand side is a
function f (x) instead
of 0, the equation becomes non-homogeneous.
Definition 1.1 – Second-Order Linear
Non-Homogeneous Differential Equations With
Variable Coefficients
A differential equation of the form: where b(x) or c(x) or both are non-constant continuous
functions and where f (x) is a non-0 continuous function, is |
Recall that while the equation is linear (in y, y',
and y''), 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. Neither do the functions b(x), c(x), and f (x).
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2. General Solution |
Similar to the case of constant-coefficients equations as
presented in Section
16.3.1 Theorem 2.1, the GS of a NHDE equals the
GS of the corresponding HDE plus a PS of the NHDE.
Theorem 2.1 – General Solutions Of The Non-Homogeneous Equations
Let: ((GS Of NHE) = (GS Of HE) + (A PS Of NHE)). |
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3. The Method Of Variation Of Parameters |
The solving of a HE with variable coefficients is discussed
in Section
16.2.2. So now we develop a procedure to find a PS of a
NHE with variable coefficients, called the method of variation of parameters.
In Section
16.1.3, we present a method to solve the first-order DE y' + p(x) y = q(x). Let's use this method to solve y' + p(x) y
= 0:
We've just shown that the GS of the HE:
y' + p(x) y = 0
is:
where u is a
function. We observe that the GS of the NHE is obtained from the GS of the
corresponding HE by replacing the
constant C by a
function u.
Let y_{1} and y_{2} be two linearly independents PSs of the second-order HE:
^{{3.1}} Problem & Solution 7
Of course y_{p} is a solution of the NHE [3.3] if Eq. [3.6] is satisfied and if:
We need only one y_{p},
hence only one u_{1} and one u_{2}; we choose the
simplest ones by setting the constants of integration for u_{1}
and u_{2} to 0. Anyway, lets' see what happens
if the constants of integration are some k_{1} and k_{2} respectively. Suppose:
u_{1} = h_{1}(x) + k_{1} and u_{2} = h_{2}(x) + k_{2}.
Then:
y_{p} = u_{1} y_{1} + u_{2} y_{2} = (h_{1}(x) + k_{1}) y_{1} + (h_{2}(x) + k_{2}) y_{2} = h_{1}(x) y_{1} + h_{2}(x) y_{2} + (k_{1}y_{1} + k_{2}y_{2}).
But k_{1}y_{1} + k_{2}y_{2} is a PS of the HE [3.1]. It follows that when we
substitute the expression for y_{p}
into the NHE [3.3], all the terms
containing k_{1} or k_{2} sum up to 0 ((k_{1} y_{1} + k_{2} y_{2})'' + b(x)(k_{1} y_{1} + k_{2} y_{2})' + c(x)(k_{1} y_{1} + k_{2} y_{2}) = 0). It
follows that choosing the
constants of integration to be any k_{1} and k_{2} yields the same PS y_{p}
of the NHE as choosing them both to be 0.
The above procedure replaces constants or parameters c_{1} and c_{2} in the GS of a
HE by variables u_{1} and u_{2} respectively
to obtain
a PS of the corresponding NHE. Hence it's called the method of variation of
parameters or the method of variation of
constants. It's also called the Lagrange method, as it's due to the
French mathematician Joseph Louis Lagrange (1736 -
1813).
We've essentially proved the following theorem.
Then: where c_{1} and c_{2} are arbitrary constants. |
a. Note that y_{h}
= c_{1} y_{1} + c_{2}_{ }y_{2} is the GS of
the HE [3.12] and remember that y_{p}
= u_{1} y_{1} + u_{2}_{ }y_{2} is a PS of the
corresponding NHE
[3.11]. A PS of the NHE is built on
2 linearly independent PSs of the corresponding HE.
b. A PS y_{p}
of the NHE is y_{p} = u_{1}y_{1} + u_{2}y_{2}, where y_{1} and y_{2} are linearly
independent PSs of the HE. The unknowns in the system
of equations obtained are the
derivatives of u_{1} and u_{2}, not u_{1} and u_{2}.
c. The method of variation of parameters first finds
a PS of a NHE, then adds it to the GS of the associated HE to obtain the GS
of the NHE.
Consider the NHE:
a. Verify that y_{1} = x and y_{2} = 1/x are linearly independent solutions of the homogeneous counterpart of the given equation.
b. Solve the given equation.
In integrating to get u_{1} and u_{2}, we choose u_{1} and u_{2} with the constants of integration equal to 0. Remember
we need only one
PS y_{p} = u_{1}y_{1} + u_{2} y_{2}, and so we
need only one u_{1} and one u_{2}. We choose the
simplest u_{1} and u_{2}, the ones with
the constants of
integration equal to 0.
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4. For Non-Homogeneous Equations With Constant Coefficients |
The procedure of the method of variation of parameters
doesn't require that the coefficients be variable functions. So the
method also applies to NHEs with constant coefficients.
Consider the constant-coefficients NHE:
y'' – y = e^{2}^{x}.
a. Solve the corresponding HE y'' – y = 0.
b. Solve the NHE by the method of undetermined coefficients.
c. Solve the NHE by the method of variation of parameters.
d. Are the answers in parts b and c the same?
For part a, we use the method of characteristic equation; see Section 16.2.1 Theorem 3.1.
Solution
d. Yes.
EOS
For constant-coefficients NHE, the method of undetermined
coefficients applies efficiently only when the right-hand side of the
constant-coefficients NHE:
y'' + by' + cy = f (x),
namely the function f (x),
is in the right form. See the table in Section
16.3.1 Part 6. When f (x)
isn't in the right form, another
method, for example the method of variation of parameters, is necessary.
Solve the DE y'' + y = tan x.
Solution
The GS of the NHE is y = c_{1} cos x + c_{2} sin x + (sin x – ln |sec x + tan x|) cos x – cos x sin x, or:
y = c_{1} cos
x + c_{2} sin x – cos x ln |sec x + tan x|).
EOS
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5. Various Types Of Equations And Various Methods |
We now summarize the types of second-order linear DEs and
the methods used to solve them that have been discussed in this
chapter. In the two tables below, we use the following abbreviations:
“coeff” |
for |
“coefficients”, |
“const” |
for |
“constant”, and |
“var” |
for |
“variable”. |
{5.1} Section
16.2.1 Theorem 3.1
{5.2} Section
16.2.2 Part 4
{5.3} Section 16.3.1 Theorem 3.1
{5.4} Section 16.3.1 Part 6
{5.5} Part 4
{5.6} Theorem 3.1
We now summarize the methods and their uses.
{5.7} Section 16.2.1 Theorem 3.1
{5.8} Section 16.2.2 Part 4 and Theorem 3.1 Of Same Section
{5.9} Section 16.3.1 Part 6
{5.10} Part 4 and Theorem 3.1
Problems & Solutions |
1. Consider the NHE:
a. Verify that y_{1} = x and y_{2} = x^{2} are linearly independent solutions of the homogeneous
counterpart of the above equation.
b. Solve the above equation.
using integration by parts we have:
2. Consider the constant-coefficients NHE:
y'' – 3y' + 2y = 6e^{3}^{x}.
a. Solve the corresponding HE y'' – 3y'
+ 2y = 0.
b. Solve the NHE by the method of undetermined coefficients.
c. Solve the NHE by the method of variation of parameters.
d. Are the answers in parts b and c the same?
The GS of the NHE is y = c_{1}e^{x} + c_{2}e^{2}^{x} – 3e^{2}^{x}e^{x} + 6e^{x}e^{2}^{x} = c_{1}e^{x} + c_{2}e^{2}^{x} + 3e^{3}^{x}.
d. Yes
3. Consider the constant-coefficients NHE:
y'' + 4y = 3 sin x.
a. Solve the corresponding HE y'' + 4y = 0.
b. Solve the NHE by the method of undetermined coefficients.
c. Solve the NHE by the method of variation of parameters.
So the GS of the NHE is y = A cos 2x + B sin 2x + sin x.
4. Find the GS of the constant-coefficients NHE:
y'' + 4y = 16x sin 2x.
See also Problem & Solution 5.
When finding a PS of a constant-coefficients NHE, if f (x) is in the right form for
the method of undetermined coefficients and if
we're not asked to use the method of variation of parameters, we should try the
method of undetermined coefficients, since it
doesn't involve integration. However for this method, we have to remember the
correct form of the working trial PS.
b_{1} = 0, a_{1} = 16/(–8) =
–2, a_{0} = –2(0)/(–4) =
0, b_{0} = –2(–2)/4 =
1,
y_{p} = ((–2)x^{2} + 0x) cos 2x + (0x^{2} + 1x) sin 2x = – 2x^{2} cos
2x + x
sin 2x.
Consequently the GS of the NHE is y = A cos 2x + B sin 2x – 2x^{2} cos 2x + x sin 2x = (– 2x^{2} + A) cos 2x + (x + B) sin 2x.
5. Prove by using the method of variation of parameters that the GS of the constant-coefficients NHE:
y'' + 4y = 16x sin 2x
is:
y = (– 2x^{2} + A) cos 2x + (x + B) sin 2x,
where A and B are arbitrary constants. See also Problem & Solution 4.
Then:
Because A + 1/4 is an arbitrary constant just like A, let's re-use the letter A for A + 1/4. Therefore the GS of the NHE is:
y = (– 2x^{2} + A) cos 2x + (x + B) sin 2x.
In Problem & Solution 4,
a PS of the equation y'' + 4y = 16x sin 2x is found to be y_{p}
= – 2x^{2} cos 2x + x sin
2x. In this Problem
& Solution 5, a PS of the same equation is found to be y_{p}
= (– 2x^{2} + 1/4) cos 2x + x sin
2x. It’s easy to verify that they both
indeed are solutions of the mentioned equation.
6. Determine the GS of the constant-coefficients NHE:
Here f (x)
isn’t in the right form for the method of undetermined coefficients. So we
won't waste our time trying that method. We
have to employ the method of variation of parameters.
So the GS of the NHE is:
y = c_{1}e^{x} + c_{2}xe^{x} – e^{x} ln |1 – x| – e^{x} = e^{x}(c_{1} + c_{2}x – ln |1 – x| – 1) = e^{x}(c_{1} + c_{2}x – ln |1 – x|),
where we re-use the letter c_{1} for the arbitray constant c_{1} – 1.
7. We now prove that the PS y_{p}
= u_{1} y_{1} + u_{2} y_{2} of the NHE y'' + b(x) y' + c(x) y
= f (x), where y_{1} and y_{2} are linearly
independent PSs of the homogeneous
counterpart of this DE, doesn't depend on the expression:
that's in Eq. [3.5] and that we set equal to 0 as done in Eq. [3.6], for which the PS is:
c. Prove that:
Solution
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