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Module 3

Differential equations

General characteristic of the module. The mathematician is generally interested in the purely formal processes and results of a mathematical investigation, independent of any particular interpretation. The physicist or engineer is likely to be interested in a result because of its meaning or interpretation when applied to some particular physical situation. The mathematical procedures, the particular interpretation, both are essential. In this chapter purely formal work and applications go hand in hand.

Topic 1.		Text organization
Topic 1.		Differential equations of the first order.
Topic 2.		Differential equations of order higher than the first.
Topic	3.	Linear differential equations with constant coefficients.
Topic	4.	Simultaneous equations (Systems of the differential equations).

Basic concepts. 1. Elementary differential equations. 2. Particular solutions (Cauchy’s problem). 3. The most general and particular solutions.

4. Simultaneous equations.

Basic problems. 1. Finding the most general solution of the differential equation. 2. Finding the particular solution of the differential equation through the given point. 3. Finding the most general solution of the simultaneous equations. 4. Compound-interest law problems.

Topic 1 .Differential equations of the first order and the first degree

Definitions and elementary problems. Solution of differential equation. Variables separable. Homogeneous equations. Linear differential equations. Equations reducible to linear form. Exact differential equation. Applications.

Literature: [2, section 3, ch.3.1], [4, section 8, §25], [6, section 11, ch.11.1], [8, section 13, §§1–9], [10, §§1–2].

T1. Main concepts

1.1. Definitions and elementary problems

An equation involving differentials or derivatives in its expression is a differential equation. If the equation contains total differentials, total

190

derivatives, or both, but does not contain partial derivatives, it is called an ordinary differential equation; if it contains partial derivatives, it is called a partial differential equation.

The order of a differential equation is the order of the highest-ordered derivative involved in its expression. For instantce, the differential equation

y xy 2y 4 0 is of the first order.

The degree of a differential equation is the degree that its highest-ordered derivative would have if the equation were rationalized and cleared of fractions with regard to all derivatives involved in it.

A solution, or integral, of an ordinary differential equation in two variables is a relation between the variables which satisfies the equation.

If, in particular, the solution has the form y = f(x), then replacement of y and the derivatives of y with respect to x by f(x) and its derivatives produces an identity.

The general solution of the first-order differential equation y f (x, y) or F(x, y, y ) 0 is a function y = j(x, C) , where C represents an arbitrary

constant.

A particular solution to the differential equation is a function y= (x, C0) such that y(x0 ) = y0 .

1.2. Variables separable

Only certain types of differential equations can be solved in terms of ordinary functions, and generally each type calls for a special method of integration.

We shall first consider a differential equation of the first order and first degree, which can easily be reduced to the form

M (x)dx N ( y)dy 0 ,

(3.1)

Where M(x) is a function of x alone and N(y) is a function of y alone. This type is referred to as variables separable. Direct integration of equation (3.1) gives the general solution

M (x)dx N ( y)dy C ,

(3.2)

Where C is an arbitrary constant. It is evident that equation (3.1) would result from differentiating equation (3.2), even if we should write any function of C instead of C.

A differential equation

M1(x)N1( y)dx M2 (x)N2 ( y)dy 0

191

is also a separable differential equation. Dividing it by N1 ( y)M 2 (x) 0 , we have

	M1 (x)	dx	N 2 ( y)	dy 0 .

	M 2 (x)		N1 ( y)
Remark. It may happen that the following solutions may be lost.
	N1 ( y)M 2 (x) 0 .				(3.3)

1.3. Homogeneous equations

A homogeneous expression of the mth degree in x and y is an expression such that if x and y are replaced by tx and ty, the result will be the original expression multiplied by tm, or, analytically expressed,

f (tx,ty) t m f (x, y) .

Thus, x2 – y2 is homogeneous in x and y for (tx)2 – (ty)2 = t2(x2 – y2). In fact, any polynomial all terms of which are the same degree in x and y is

homogeneous. It appears also that such an expression as

x2 y2

y sin

is homogeneous, since

ty 3

ty sin

y sin

= t

3 .

The function

f (x, y)

x2 y 2

is homogeneous expression of the zero

2x2 y 2

degree, therefore

f (tx, ty)

t 2 x2 t 2 y 2

t 2 (x2 y 2 )

t 0

f (x, y) .

2t 2 x2 t 2 y 2

t 2 (2x2 y 2 )

A differential equation

y f (x, y)

is homogeneous in x and y if f (x, y)

is homogeneous function of the zero degree.

A differential equation

M (x, y)dx N (x, y)dy 0

is homogeneous

in x

and

y if M (x, y)

and

N(x, y)

are

homogeneous

functions of the same degree in x and y.

To solve for y , we convert the homogeneous differential equation to the form:

y g y .x

192

Either of the substitutions, y = xu(x), where u(x) is the new unknown function, will reduce any homogeneous differential equation of the first order and first degree to the type of variables separable.

For the equations of the form



y			a1x b1 y c1

		f
		a2 x b2 y c2
we could have used the substitutions				x t ,	y v , where t		and v are
the new variables, α and β are solutions of the system
a		α b		β c 0,			(3.4)
	1	1		1
a2 α b2 β c2 0
such circumstances a1b2 a2b1 0 . If						0 , the	substitution

z a1x b1 y c1 transforms the given expression to homogeneous differential

equation.

1.4. Linear differential equation. Equations reducible to linear form

A differential equation M (x, y)dx N (x, y)dy 0 of any order is said to

be linear when it is of the first degree in the dependent variable and its derivatives. It follows that the general type of a linear differential equation of the first order is

y P(x) y Q(x),

(3.5)

where P and Q are functions of x only.

To find an integrating factor of (3.5) let us solve

y P(x) y 0 .

(3.6)

Here the variables are separated and the solution is

P(x) y ;

dy P(x)dx ;

P(x)dx ;

Finally,

P(x)dx ln

y Ce

P(x)dx

Thereafter, to solve an equation having the form (3.5) two following methods by Bernoulli and by Lagrange are used.

Bernoulli’s method. Solution of (3.5) is defined as the product of the two unknown functions u(x) and v(x) , namely

y(x) u(x)v(x)

193

Substituting in (3.5), we get

du v u dv P(x)uv Q(x) ,
dx			dx
or
du					dv	Q(x).
		P(x)u v u			dx	Q(x).
dx					dx
Define u(x) such as the expression in a parenthesis be equal to zero, then
	du		P(x)u 0,
			P(x)u 0,
	dx					(3.7)
		dv		Q(x).
	u			Q(x).
		dx
		dx

The first equation has the following particular solution (C = 0):

u(x) e

P(x)dx

The second equation of (3.7) has the form (3.6), hence

dv -ò P(x)dx

ò P(x)dx

, v Q(x)e

P(x)dx

dx e

=Q(x),

=Q(x)e

dx C.

Thus

-ò P(x)dx

ò Q(x)e

ò P(x)dx

y =uv =Ce

dx.

(3.8)

Lagrange’s method. Solution of (3.5) is defined in the form

y C(x)e

P(x)dx

(3.9)

This has the form (3.6) if C is a constant. If C(x) is a new function substituting

		P(x)dx		. Therefore,
(3.9) in (3.5), we get C (x) Q(x)e				. Therefore,
C(x) = ò	Q(x)e		ò	P(x)dx	dx +C1	,
C(x) = ò	Q(x)e				dx +C1	,
where C1 is any constant. Then, the general solution is
y = e-ò P(x)dx	é					ù
y = e-ò P(x)dx	êêC1 +		ò	Q(x)eò P( x)dx dxúú ,
	ê					ú
	ë					û

which is identity with (3.8).

Remark. Sometimes the differential equation is easily computed if x is a function of y, namely x P( y)x Q( y) .

Equations reducible to linear form. The equation

194

y P(x) y Q(x) yk ,

where k is a real number (k 0; 1 ), is named Bernoulli’s equation. To solve

it, we first divide by yk and obtain

y-k y + P y1-k = Q. This form suggests the substitution

z y1 k

, z = (1 – k) y-k y .

Substituting, we get

	z	Pz Q ,
1 k		Pz Q ,
1 k
a linear equation.

1.5. Exact differential equation

The formula for the total derivative of function u(x, y) is du ux dx uy dy M (x, y)dx N (x, y)dy .

Any expression which is exactly the total differential of some function of x and y is called an exact differential, and such an expression equated to zero is an exact differential equation. It appears then that an expression

M (x, y)dx N (x, y)dy 0

is exact if there exists a function u(x, y)

such that

M = u

N =

u .

From this we obtain

y x

x y

Since

, it appears that

y x

x y

M Ny x

is necessary condition that the left side of (3.10) be an exact differential. If M = ux is hold,

(3.10)

(3.11)

195

u = x M dx + (y),	(3.12)
x0

where (y) does not contain x and the superscript x means that y is to be held constant during the integration. Substituting the value of u from (3.12) in

N= uy , we obtain

u		x
y			M dx			dy	N ,
y		y	M dx			dy	N ,
or			x0
or				x
				x
	y	N			M dx.
	y	N		y	M dx.
					x0

Since the partial derivative with respect to x of the right member is N M ,

x y

it is zero. Therefore the right member does not contain x. Hence we find, by integrating, that is

	y			x		y
							N x0 , y dy C.
=		N			M dx dy
=		N	y		M dx dy
			y
	y0			x0		y0
Substituting in (3.12) we get
		x			y
u(x, y) = M x, y dx N x0 , y dy C = 0.
		x0			y0

1.6. Simple applications

Any differential equation of the first degree may be written in the form

M(x, y)dx + N(x, y)dy = 0, or y = – M x, y . (3.14)

N x, y

As y evaluated for a curve represents its slope, it appears that equation (3.14) assigns to every point in the plane of a slope and that any curve represented by a solution of (3.14) must have a slope given by (3.14). If, then, we can find the expression for the slope of a curve, we can find the curve as a solution of a differential equation. If the equation of a particular curve passing through a point is wanted, the constant of integration must be replaced by such a number that the coordinates of the point will satisfy the equation.

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