Ly = h(x) The Nonhomogeneous Equation

Chapter VI: Boundary Value and Eigenvalue Problems

II. Ly = h(x) The Nonhomogeneous Equation

Solutions to the nonhomogeneous equation can be obtained with the help of an ansatz that goes back to Lagrange, the method of variation of constants. In this method, the constant C in the general solution y(x; C) = of the homogeneous equation is replaced by a function C(x). The calculation of an appropriate choice of C(x) gives a solution of the nonhomogeneous equation.

Indeed, the ansatz

y(x) = with G(x)

=

f

^{g(t) dt}

leads to

Ly y' + gy = (C' —gC+ =

Hence Ly = h holds if and only if

C' =

or equivalently, C(x)

=

f

h(t)eG(t) dt + C0. (5)

Theorem. If the fanctions g(x), h(x) are continuous in J and J, then the initial value problem

Ly = ^y'+ g(x)y = h(x), = ⁽⁶⁾

has exactly one solution,

y(x) = ^. +^e_C(z)

f

^{dt. (7)}

The solution exists in all of J.

The discussion leading up to formula (5) shows that (7) is a solution to

Ly =

^h; it is clear that the initial conditions are satisfied. Uniqueness is a consequence of (a) below.

Remark on Linearity. If y, are two solutions to the nonhomogeneous equation Ly = h, then L(y —

= Ly

^— Ly = ^0,

i.e., z(x) =

y —

is a

§ 2. The Linear Differential Equations. Related Equations 29

solution of the homogeneous equation Ly = 0. Thus all solutions y(x) of the nonhomogeneous equation can be written in the form

y(x) = + z(x), (8)

where is a fixed solution of the nonhomogeneous equation and z(x) runs through all solutions of the homogeneous equation. In other words,

y(x; C) = ^{+ Ce_C(s)} (C E IR) (8')

is the general solution of the inhomogeneous equation.

It follows from (4) that a solution z of the homogeneous equation that van- ishes at a point is identically zero (note that e canbe any point in J). Using (8), this result implies

(a) Two solutions y, of the inhomogeneous equation that coincide at one point in J are identical.

Example.

y' + y sin x = sin3x.

Here G(x) = ^{— cosx.} Hence z(x; C) = ^Cecoss is the general solution of the homogeneous equation Lz =⁰ and

V(x) =

f

^{sin3 t .eCOS5—C0St dt}

cos 5

=ecOSS

f

^{— 1)e8ds}

cos S

= _ec0s5((s2_1)+28+2)e_8

=sin2x —2cosx —2 +^4ec0s5_l

is a solution to the nonhomogeneous differential equation. It follows that the general solution of the nonhomogeneous equation is given by

y(x; C) =sin2x —2cosx—2 + C• eCOSS.

iii.

^g(x)y+ h(x)ya = ^0,

a

1

Bernoulli's Equation.

This differential equation, named after Jacob Bernoulli (1654—1705), can be transformed into a linear differential equation. Let us assume that the functions

g, h are continuous in J and that y > 0.

If the equation is multiplied by

(1 — and the relation (1 — = is used, then one obtains (yl_a)I +^{(1 — a)g(x)yl_a+ (1 —a)h(x)}

=

^0.

Thus the function z = satisfies a linear differential equation,

z' + (1 — a)g(x)z+ (1 —a)h(x) = ^0. (9)

y

Solution of the initial value problem

V+

_j-f— ^{+(1+x)y4 =0, y(O) = —1}

Conversely, if z(x) is a positive solution of (9), then the function y(x) =

(z(x)) isa positive solution of Bernoulli's differential equation. For > 0,

the initial condition

=

transforms into z(e) = > 0. By Theorem II, this condition uniquely defines a solution z of (9). Hence each initial value problem for the Bernoulli equation with a positive initial value at the point is uniquely solvable.

The cases where also nonpositive solutions occur will be discussed now.

(a) a >

^0: Then the differential equation is defined for y 0, and y 0

is a solution. Since all positive solutions can be given explicitly, it is easy to determine, on a case by case basis, whether or not solution curves run into the x-axis from above. This is the case, for example, for g = 0,

h =

^—1,

a

(Example 2 from i.v).

(b) a an integer: Then y <⁰ is also permitted. There are two cases.

a odd: It follows from the Bernouffi equation that

(—y)' +g(x)(—y) + = 0.

So if y(x) is a positive solution of the Bernoulli equation, then u(x) = —y(x) is a negative solution. Hence initial value problems with ii < 0 can be easily handled.

a even: Since 1 — a is odd, y < ⁰ implies z = <0, which in turn yields

= So for a negative initial value thenegative solution z of (9) with z(e) = leadsto a negative solution ii = with =

Inboth cases the solution y satisfying = <0 is unique.

Exercise. Show directly (without using the uniqueness theorem) that for a 2, a E N, a solution y 0 of Bernoulli's equation has no zero in J.

Example.

+(1+x)y4

=0.

§ 2. The Linear Differential Equations. Related Equations 31

The differential equation is defined for both positive and negative y. Using

z gives, according to (9), y

z'— =0.

1+x

Clearly, q5 = C(1 + x)3 is the general solution of the homogeneous equation.

Thus the ansatz for the nonhomogeneous equation (by variation of constants) is z =C(x)(1+ x)3. After a simple calculation, one obtains

/ 3

C

Therefore, the general solution of the nonhomogeneous equation is

z(x;C)=C(1+x)3—3(1+x)2=(1+x2)(Cx+C--3).

Since a = ⁴is even, one has

sgn(Cx+C—3)

The solution through the point (0, —1) is given by

IV. y' + g(x)y + h(x)y2 = ^k(x)

Riccati's Equation.

In this equation, which is named after the Italian mathematician ^Jacopo

Francesco Riccati (1676—1754), the functions g(x), h(x), k(x) are assumed to be continuous in an interval J. Except in special instances, the solutions cannot be given in closed form. However, if one solution is known, then the remaining solutions can be explicitly calculated. For proof, we consider the difference of

two solutions y and q5, u(x) =y(x) ^— it satisfies the equation

u'+gu+h(y2

= 0.

Since y2 — = (y^— + = u(u+ ^{one has}

= [g(x)+ 2çb(x)h(x)]u + h(x)u2 =^{0. (10)}

Thus the difference satisfies a Bernoulli differential equation which can be con- verted, using the techniques described in III, into the linear differential equation

— [g(x) + = h(x), where z(x)

=

⁽¹¹⁾

Summary. If a solution q5(x) of the Riccati equation is known, then all of the other solutions can be obtained in the form

(12) where z(x) is an arbitrary solution of the linear equation (11).

Example. — — 2xy

=

2.

The function 4(x) = is a particular solution. Formula (11) then gives the linear differential equation

2x——

+1=0.

"

xl

The general solution to the homogeneous equation is z(x) = from which a particular solution 2 of the nonhomogeneous equation can be obtained using (7) with h =—1:

2(x) =^_x2e_x2 ex2

J

^{— dx}

=_x2e_x2 (_1ex2 _{+ 2}

feZ2 dx)

— with E(x)

=

f

The integral E(x) can be expressed in terms of the error function with imaginary argument. The general solution of the original Riccati equation is now obtained from (12),

1 1

y(x; C) = +

x + ^— 2E(x))

—2E(x))

— 1+ (C — 2E(x))

Since y(O; C) = C, every initial value problem y(O) = can be immediately solved.

V. Exercises. (a) Isoclines. Isoclines of a differential equation y' = f(x,y) are the curves f(x, y) = const, on which the direction field has constant slope. Sketch the direction field for the differential equation

=

^{y2+ 1— x2}

making use of the isoclines y2 + 1 —x2 = const. Determine all solutions (one solution is evident from the direction field). Which solutions exist on an infinite interval; which exist in R?

(b) Determine all solutions of the differential equations

!/+ysinx=sin2x and y'—3ytanx=1.

(c) Solve the initial value problem

= x4y+ x4y4, y(O) =

§ 2. The Linear Differential Equations. Related Equations 33

VI. Exercise. Suppose f(s) is continuous on the half-open interval 0 <

x < 1. What additional conditions must f(s) satisfy so that every solution of the differential equation

y'—f(x)y for 0<xl

has the property

(a) y(x)—÷0

as x—*+0?

(b)

Investigate the same question for the differential equation for

0 y(x) l/e are taken into consideration.

VII.

Exercise. The Riccati Differential Equation and Linear Differen- tial Equations of Second Order. Show that the Riccati differential equation

y'+g(x)y+h(x)y2 =

^k(s)

with g,h E C°(J), h E C'(J), h(s) 0 in J, can be transformed into the linear differential equation of second order

(13)

using the transformation

u(s)

exp (f h(x)y(x)

dx)

and that conversely, a positive solution u of (13) produces a solution y =

(logu)'/h of the Riccati equation. Use this relationship to solve the initial value problem

— +

^{exy2 + =0, y(O) =}

Supplement: The Generalized Logistic Equation.

We consider a generalization of the logistic differential equation ii' = u(b^— cu),where, in contrast to 1.XIII, b and c depend on t. Our objective is to derive some theorems on the asymptotic behavior of the solutions as t oo and on the existence of a class of distinctive, in particular periodic, solutions.