Advanced Mathematical Methods for Scientists and Engineers Episode 3 Part 8 ppsx

Any function yp which satisfies this equation is called a particular solution of the differential equation.. For somesimple differential equations, primarily constant coefficient equatio

Now consider the case that y0(x) is negative on the interval so y(a) > y(b).

Z b a

δ(y(x)) dx =

Z y(b) y(a)

δ(y) dydx

−1

dy

=

Z y(b) y(a)

δ(y)

y0(x)dy

=

Z y(a) y(b)

δ(y)

−y0(x)dy

=

(Ry(a) y(b)

δ(y(x)) dx =

(Rβ α

Now we turn to the integral of δ(y(x)) on (−∞ ∞) Let αm = min(y(ξm), y(ξm)) and βm = max(y(ξm), y(ξm)).

we will use integration by parts.

The Dirac delta function is defined by the following two properties

δ(x − a) = 0 for x 6= aZ

δ(ξ − α)

|J| |J| dξ = δ(ξ − α) dξ

=

Zδ(ξ1 − α1) · · · δ(ξn− αn) dξ

=

Zδ(ξ1 − α1) dξ1· · ·

Zδ(ξn− αn) dξn

Z 2π 0

Z ∞ 0

δ3(x − x0) r2sin(φ) dr dθ dφ = 1For r0 6= 0, and φ0 6= 0, π, the Dirac Delta function is

δ3(x − x0) = 1

r2sin(φ)δ (r − r0) δ (θ − θ0) δ (φ − φ0)since it satisfies the two defining properties

δ (r − r0) dr

Z 2π 0

δ (θ − θ0) dθ

Z π 0

δ (φ − φ0) dφ = 1

For φ0 = 0 or φ0 = π, the Dirac delta function is

2

sin(φ) dr dθ dφ

= 12π

Z ∞ 0

δ (r − r0) dr

Z 2π 0

dθ

Z π 0

δ (φ − φ0) dφ = 1For r0 = 0 the Dirac delta function is

Z ∞ 0

14πr2δ (r − r0) r2sin(φ) dr dθ dφ = 1

4π

Z ∞ 0

δ (r) dr

Z 2π 0

dθ

Z π 0

sin(φ) dφ = 1

Chapter 21

Inhomogeneous Differential Equations

Feelin’ stupid? I know I am!

-Homer Simpson

Consider the nth order linear homogeneous equation

L[y] ≡ y(n)+ pn−1(x)y(n−1)+ · · · + p1(x)y0+ p0(x)y = 0

Let {y1, y2, , yn} be a set of linearly independent homogeneous solutions, L[yk] = 0 We know that the generalsolution of the homogeneous equation is a linear combination of the homogeneous solutions

L[y] ≡ y(n)+ pn−1(x)y(n−1)+ · · · + p1(x)y0+ p0(x)y = f (x)

Any function yp which satisfies this equation is called a particular solution of the differential equation We want toknow the general solution of the inhomogeneous equation Later in this chapter we will cover methods of constructingthis solution; now we consider the form of the solution.

Let yp be a particular solution Note that yp+ h is a particular solution if h satisfies the homogeneous equation

L[yp+ h] = L[yp] + L[h] = f + 0 = fTherefore yp+ yh satisfies the homogeneous equation We show that this is the general solution of the inhomogeneousequation Let yp and ηp both be solutions of the inhomogeneous equation L[y] = f The difference of yp and ηp is ahomogeneous solution

L[yp− ηp] = L[yp] − L[ηp] = f − f = 0

yp and ηp differ by a linear combination of the homogeneous solutions {yk} Therefore the general solution of L[y] = f

is the sum of any particular solution yp and the general homogeneous solution yh

Example 21.1.1 The differential equation

y00+ y = sin(2x)has the two homogeneous solutions

y1 = cos x, y2 = sin x,

and a particular solution

cosx2



is a particular solution

The first method we present for computing particular solutions is the method of undetermined coefficients For somesimple differential equations, (primarily constant coefficient equations), and some simple inhomogeneities we are able

to guess the form of a particular solution This form will contain some unknown parameters We substitute this forminto the differential equation to determine the parameters and thus determine a particular solution

Later in this chapter we will present general methods which work for any linear differential equation and anyinhogeneity Thus one might wonder why I would present a method that works only for some simple problems (Andwhy it is called a “method” if it amounts to no more than guessing.) The answer is that guessing an answer is lessgrungy than computing it with the formulas we will develop later Also, the process of this guessing is not random,there is rhyme and reason to it

Consider an nth order constant coefficient, inhomogeneous equation

L[y] ≡ y(n)+ an−1y(n−1)+ · · · + a1y0+ a0y = f (x)

If f (x) is one of a few simple forms, then we can guess the form of a particular solution Below we enumerate somecases

f = p(x) If f is an mth order polynomial, f (x) = pmxm+ · · · + p1x + p0, then guess

yp = cmxm+ · · · c1x + c0

f = p(x) eax If f is a polynomial times an exponential then guess

yp = (cmxm+ · · · c1x + c0) eax

f = p(x) eaxcos (bx) If f is a cosine or sine times a polynomial and perhaps an exponential, f (x) = p(x) eaxcos(bx)

or f (x) = p(x) eaxsin(bx) then guess

yp = (cmxm+ · · · c1x + c0) eaxcos(bx) + (dmxm+ · · · d1x + d0) eaxsin(bx)

Likewise for hyperbolic sines and hyperbolic cosines

Example 21.2.1 Consider

y00− 2y0+ y = t2.The homogeneous solutions are y1 = et and y2 = t et We guess a particular solution of the form

yp = at2+ bt + c

We substitute the expression into the differential equation and equate coefficients of powers of t to determine theparameters

y00p − 2yp0 + yp = t2(2a) − 2(2at + b) + (at2+ bt + c) = t2(a − 1)t2+ (b − 4a)t + (2a − 2b + c) = 0

a − 1 = 0, b − 4a = 0, 2a − 2b + c = 0

a = 1, b = 4, c = 6

A particular solution is

yp = t2+ 4t + 6

If the inhomogeneity is a sum of terms, L[y] = f ≡ f1+· · ·+fk, you can solve the problems L[y] = f1, , L[y] = fkindependently and then take the sum of the solutions as a particular solution of L[y] = f

yp = at2et

We substitute the expression into the differential equation and equate coefficients of like terms to determine theparameters.

yp00− 2yp0 + yp = et(at2+ 4at + 2a) et−2(at2+ 2at) et+at2et= et

2a et = et

a = 12

a = 110

A particular solution is

yp = x

3

10.

21.3 Variation of Parameters

In this section we present a method for computing a particular solution of an inhomogeneous equation given that weknow the homogeneous solutions We will first consider second order equations and then generalize the result for nth

order equations

Consider the second order inhomogeneous equation,

L[y] ≡ y00+ p(x)y0+ q(x)y = f (x), on a < x < b

We assume that the coefficient functions in the differential equation are continuous on [a b] Let y1(x) and y2(x)

be two linearly independent solutions to the homogeneous equation Since the Wronskian,

W (x) = exp



−

Zp(x) dx

,

is non-vanishing, we know that these solutions exist We seek a particular solution of the form,

yp = u1(x)y1+ u2(x)y2

We compute the derivatives of yp

yp0 = u01y1+ u1y01+ u02y2+ u2y02

y00p = u001y1+ 2u01y01+ u1y100+ u002y2+ 2u02y20 + u2y002

We substitute the expression for yp and its derivatives into the inhomogeneous equation and use the fact that y1 and

y2 are homogeneous solutions to simplify the equation

u001y1+ 2u01y10 + u1y100+ u002y2 + 2u02y02+ u2y200+ p(u01y1+ u1y01+ u02y2+ u2y20) + q(u1y1+ u2y2) = f

u001y1+ 2u01y10 + u002y2+ 2u02y20 + p(u01y1 + u02y2) = f

This is an ugly equation for u1 and u2, however, we have an ace up our sleeve Since u1 and u2 are undeterminedfunctions of x, we are free to impose a constraint We choose this constraint to simplify the algebra.

u01y1+ u02y2 = 0This constraint simplifies the derivatives of yp,

u02



= 0f

Here W (x) is the Wronskian.

W (x) =

y1 y2

y01 y20

We integrate to get u1 and u2 This gives us a particular solution

Result 21.3.1 Let y1 and y2 be linearly independent homogeneous solutions of

L[y] = y00 + p(x)y0+ q(x)y = f (x).

where W (x) is the Wronskian of y1 and y2.

Example 21.3.1 Consider the equation,

y00+ y = cos(2x)

The homogeneous solutions are y1 = cos x and y2 = sin x We compute the Wronskian

W (x) =

... is

yp = u1y1 + u2y2 + · · · + unyn where... data-page="27">

Result 21.5 .3 The problem

L[y] = y(n)+ pn−1y(n−1)+ · · · + p1y0+ p0y... where

L[u] = f (x), B1[u] = 0, B2[u] = 0, and L[v] = 0, B1[v] = γ1, B2[v] = γ2.