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Undetermined Coefficients and Cauchy-Euler

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Higher-Order Differential

Equations
Linear Nonhomogeneous Differential Equations with Constant
Coefficients
Nonhomogeneous Linear DE
To solve a nonhomogeneous linear differential equation,

a
ny(
n )
a
n
1y(
n 1)
 
...
a y
1 '
a0yg
(x)
we must do two things:
1.) Find the complementary function yc
2.) Find any particular solution yp of the nonhomogeneous equation.

The general solution of the nonhomogeneous linear DE on an interval I


is y = yc + yp
Nonhomogeneous Linear DE
The complementary function yc is the general solution of the
associated homogeneous DE of the original nonhomogeneous
differential equation, that is

an y ( n )  an1 y ( n1)  ...  a1 y ' a0 y  0


Method of Undetermined Coefficients
The underlying idea in this method is a conjecture, an educated guess
really, about the form of yp motivated by the kinds of functions that
make up the input function g(x). The general method is limited to non-

homogeneous linear DEs such as n a y(n)
an1y (
n 1)
 
...a1y'a0 y g (x)
where:
• the coefficients a0, a1, ..., an are constants, and
• g(x) is either a constant, a polynomial function, an exponential
function eαx, a sine function sin βx, a cosine function cos βx, or finite
sums and products of these functions
Method of Undetermined Coefficients
Method of Undetermined Coefficients
Case 1
No function in the assumed particular solution is a solution of
the associated homogeneous differential equation.
Example:
y”-8y'+25y = 5x3e-x - 7e-x

y”-8y'+25y = (5x3 - 7)e-x


yc = e4x (c1cos 3x + c2sin 3x)
yp = (Ax3 + Bx2 + Cx + E)e-x
No duplication between terms in yp and the terms in yc
Method of Undetermined Coefficients
Case 2
A function in the assumed particular solution is also a solution of
the associated homogeneous differential equation.
Example:
y”-5y'+4y = 8ex

yc = c1ex + c2e4x
yp = Aex
Substituting to the differential equation yields 0 = 8ex. Use yp = Axex
Use yp = Axn+1eαx for yc containing cxneαx to eliminate all duplications.
Example 1 (Undetermined Coeffieients Case 1)
Solve y” - 2y' - 3y = 4x - 5 + 6xe2x

Ans: y = c1e-x + c2e3x - 4x/3 + 23/9 - (2x + 4/3)e2x


Example 2 (Undetermined Coefficients Case 2)
Solve the IVP y” + y = 4x + 10 sin x , y(π) = 0, y'(π) = 2

Ans: y = 9π cos x + 7 sin x + 4x - 5x cos x


Cauchy-Euler Equations
Any linear differential equation of the form
an x n y ( n )  an1 x n1 y ( n1)  ...  a1 xy ' a0 y  g ( x)
where the coefficients an, an-1, ..., a0 are constants, is known as a
Cauchy-Euler equation in honor of and
. The observable characteristic of this type of equation
is that the degree k = n, n-1, ..., 1, 0 of the monomial coefficients xk
matches the order of differentiation dky/dxk.
Cauchy-Euler Equations
Note:
The coefficient of dny/dxn is zero at x=0. Hence, we confine our
attention to finding the general solution on the interval (0,∞). Solutions
on the interval (-∞,0) can be obtained by substituting a variable t = -x
into the differential equation.
Method of Solution
We try a solution of the form y = xm where m is to be determined.
Analogous to what happened when we substitute emx into a linear
equation with constant coefficients, after substituting xm, each term of
a Cauchy-Euler equation becomes a polynomial in m times xm. For
example, by substituting y = xm to a second-order DE, it becomes
y  xm
y '  mxm1
y ' '  m(m  1) x m2
a2 x 2 y ' ' a1xy ' a0 y  a2 x 2 m(m  1) x m2  a1 xmxm1  a0 x m
 a2 m(m  1) x m  a1mxm  a0 x m
Method of Solution
Thus y = xm is a solution of the differential equation whenever m is a
solution of the auxillary equation a2m(m-1) + a1m + a0 = 0.
Case 1. Distinct Real Roots (m1 ≠ m2)
y = c1xm1 + c2xm2

Case 2. Repeated Real Roots (m1 = m2)


y = c1xm1 + c2xm1 ln x

Case 3. Conjugate Complex Roots (m = α ± βi)


y = xα [c1 cos(β ln x) + c2 sin((β ln x)]
Example 3 (Cauchy-Euler Case 1)
Solve: x2y'' - 2xy' - 4y = 0

Ans: y = c1x-1 + c2x4


Example 4 (Cauchy-Euler Case 2)
Solve: 4x2y'' + 8xy' + y = 0

Ans: y = c1x-1/2 + c2x-1/2 ln x


Example 5 (Cauchy-Euler Case 3)
Solve the IVP: 4x2y'' + 17y = 0, y(1) = -1, y'(1) = -1/2

Ans: y = -x1/2 cos (2 ln x)

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