1 Method for Ordinary Differential Equations

This chapter will introduce the reader to the terminology and notation of differential equations.Students will also be reminded of some of the elementary solution methods they are assumed tohave encountered in an undergraduate course on the subject. At the conclusion of this review oneshould have an idea of what it means to ‘solve’ a differential equation and some confidence thatthey could construct a solution to some simple and special types of differential equations.

1.1 Definitions and Notation

Differential equations are divided into two classes, ordinary and partial. Anordinary differentialequation (ODE) involves one independent variable and derivatives with respect to that variable. Apartial differential equation (PDE) involves more than one independent variable and correspondingpartial derivatives. The first semester of this course is about ordinary differential equations. Thephrase ‘differential equation’ will usually mean ordinary differential equation unless the contextmakes it clear that a partial differential equation is intended. The order of the highest derivative inthe differential equation is theorder of the equation.

Example (1.1.1)

(a) x2 d3y

dx3− 6x

dy

dx+ 10y = 0 3rd order ODE

(b) x′′(t) + sinx(t) = 0 2nd order ODE

(c) y′′ + λy = 0 2nd order ODE

(d) x′ + x = 0 1st order ODE

(e)∂u

∂x− ∂u

∂y= 0 1st order PDE

(f)∂2u

∂t2=

∂2u

∂x2+∂2u

∂y22nd order PDE

From the equation it is usually clear which is the independent variable and which is dependent.For example, in (a) it is implied thaty = y(x) while in (b) the dependence ofx upont is explicitlynoted. In (c) and (d) the choice of the independent variable is arbitrary but as a matter of choicewe will usually regardy = y(x) andx = x(t). The latter notation is usually reserved for thediscussion of dynamical systems in which case one should think ofx(t) as denoting the state ofa system at timet. In contrast the notationy = y(x) might be used in the context of a boundaryvalue problem whenx has the interpretation of a spatial coordinate. Of course this is much ado

1

about nothing and the student should be prepared to see differential equations written in a varietyof ways.

For the partial differential equation in (e) u = u(x, y) while in (f), u = u(x, y, t). We willoften denote partial derivatives by subscripts in which case (e) could be written ux − uy = 0 andsimilarly for (f) utt = uxx + uyy.

Differential equations are divided into two other classes, linear and nonlinear. An nth orderlinear differential equation can be put into the form

a0y(n) + a1y

(n−1) + · · ·+ an−1y′ + any = b

where b and the coefficients ak depend on at most the independent variable. If b = 0 the equationis said to be homogeneous, otherwise it is called nohomogeneous.

Example (1.1.2)y′′ = y′ + 1 linear, nonhomogeneous

x′ sin t = x linear, homogeneous

(y′)2 = x + y nonlinear

General differential equations of order 1 and 2 may be written as

F (x, y, y′) = 0 (1.1.1)

F (x, y, y′, y′′) = 0

with the obvious notation for higher order equations. Such equations are often assumed to besolvable for the highest derivative and then written as

y′ = f(x, y) (1.1.2)

y′′ = f(x, y, y′).

A differentiable function y = φ(x) is a solution of (1.1.1) on an interval J ifF (x, φ(x), φ′(x)) =0, x ∈ J. Usually a first order equation has a family of solutions y = φ(x, c) depending ona single parameter c. A second order equation usually has a two-parameter family of solutionsy = ϕ(x, c1, c2) and so on. These parameters are like constants of integration and may be deter-mined by specifying initial conditions corresponding to some ‘ time’ when a process starts.

Example (1.1.3) A two-parameter family of solutions of x′′ = 6 is

x = 3t2 + c1t+ c2 (1.1.3)

The solution that satisfies the initial conditions

x(0) = 1, x′(0) = −1

isx = 3t2 − t+ 1.

2

Because solutions may involve one or more integrations, the graph of a solution is called anintegral curve. The integral curves in the previous example are parabolas. A family of functions isa complete solution of a differential equation if every member of the family satisfies the differentialequation and every solution of the differential equation is a member of the family. The family offunctions given by (1.1.3) is the complete solution of y′′ = 6.

For the student who is familiar with the terminology of a general solution we should state atthis point that the notion of a general solution will only be used in the context of linear equations.The appropriate definitions will be introduced in Chapter 3 at which time the distinction betweena general and complete solution will be made clear.

Example (1.1.4) The family

y = ϕ(x, c) =1√

2x + c

solves y′ + y3 = 0. The family is not complete since the trivial solution, y(x) = 0 (also denotedy ≡ 0), cannot be obtained for any value of c.

In equation (1.1.2) the left hand member dy/dx denotes the slope of a solution curve whereasthe right hand member f(x, y) gives the value of the slope at (x, y). From this point of viewthe solution of an ordinary differential equations is a curve that flows along directions specified byf(x, y). The use of dy/dx breaks down when the tangent is vertical but the geometric interpretationremains meaningful. To deal with this matter note that a smooth curve in the (x, y) plane may bedescribed locally by y = ϕ(x) or x = ψ(y). If dy/dx is meaningless, dx/dy may be permissible.Consequently, it may be advantageous to treat x or y as independent or dependent variables informulating a differential equation. These remarks motivate the following discussion.

A differential equation is said to be in differential form if it is written as

M(x, y)dx+N(x, y)dy = 0. (1.1.4)

A differentiable function y = φ(x) is a solution of (1.1.4) on an interval if the substitutions

y = φ(x), dy = φ′(x)dx

make (1.1.4) true. A differentiable function x = ψ(y) is a solution if the substitutions

x = ψ(y), dx = ψ′(y)dy

make (1.1.4) true. An equation

G(x, y) = c, c constant (1.1.5)

is said to furnish an implicit solution of (1.1.2) if (1.1.5) can be solved for y = φ(x) or x = ψ(y)in a neighborhood of a point (x, y) satisfying (1.1.5) and φ or ψ is a solution in the sense definedabove. When (1.1.5) furnishes an implicit solution the graph of this equation is also called anintegral curve.

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There is some ambiguity as to what one is expected to produce when asked to solve a dif-ferential equation. Often the differential equation is regarded as solved when one has derived anequation such as (1.1.5). However, the context of the problem may make it clear that one needs toproduce an explicit formula for the solution, say y = φ(x) or x = ψ(y). The next two exampleswill expound upon this issue, but in general the specific problem at hand will usually dictate as towhat is meant by the phrase, ‘solution’ .Example (1.1.5) Consider the equation in differential form

xdx + ydy = 0.

This equation is easily solved by writing it as a ‘pure derivative’ , i.e.,

1

2d(x2 + y2) = 0.

When written in this form it is apparent that possible implicit solutions are

x2 + y2 = C. (1.1.6)

If C < 0, the locus is empty. If C = 0 the locus consists of the point (0,0). This case does notcorrespond to a solution since it does not describe a differentiable curve of the form y = ϕ(x) orx = ψ(y). If C > 0, the integral curves are circles centered at the origin.

Example (1.1.6) Consider the three equations

(a)dy

dx= −x

y(b)

dx

dy= −y

x(c) xdx + ydy = 0.

(a) This form of the equation implicitly requires a solution y = φ(x) and no integral curve cancontain a point (x, y) where y = 0.

(b) This requires a solution y = ϕ(x) and no integral curve can contain a point (x, y) wherex = 0.

(c) This equation allows for solutions of the form x = ψ(y), or y = ϕ(x) and leads to thefamily of circles (1.1.6) obtained in the previous example. In particular, in the neighborhood ofany point (x, y) on the circle we can express y as a differentiable function of x or vice versa. Onecould obtain from (1.1.6) additional functions y = ϕ(x) by taking y > 0 on part of the circle andy < 0 on the remainder (as in (c) below). Such a function is not differentiable and hence not asolution.

(a) (b) (c)y=φ(x) x=ψ(y) y=χ(x)

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Fig. 1.1.1. Solutions defined by an integral curve are displayed in (a) and (b). Though the graph ofχ(x) lies on an integral curve, it is not a solution of the differential equation.

Having introduced the notion of a solution and integral curve, we now will review some of themost elementary methods for solving ordinary differential equations.

1.2 Examples of Explicit Solution Techniques

(a) Separable Equations.A differential equation is separable if it can be wrtitten in the form

F (x, y, y′) =dy

dx− f(x)/g(y) = 0.

The differential equation is solved by ‘separating’ the variables and performing the integra-tions ∫

g(y)dy =

∫f(x)dx.

Example (1.2.1) Solvedy/dx = y2/x, x �= 0.

We separate the variables and integrate to obtain∫1

y2dy =

∫1

xdx

or

y = − 1

ln |x|+ C(1.2.1)

In the above calculation division by y2 assumed y �= 0. The function y = 0 is also a solution,referred to as a singular solution. Therefore, the family given in (1.2.1) is not complete.

(b) Homogeneous Equations.A differential equation is homogeneous if it can put in the form

F (x, y, y′) = y′ − f(y/x) = 0.

In this case let z = y/x sody

dx= z +

dz

dx= f(z)

ordz

dx=

f(z)− z

x,

which is separable.

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(c) Exact Equations.The differential equation F (x, y, y′) = 0 is exact if it can be written in differential form

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

where∂M

∂y=

∂N

∂x.

Recall that if M,N are C1 on a simply connected domain then there exists a function Psuch that

∂P

∂x= M,

∂P

∂y= N.

Thus the exact differential equation may be written as a pure derivative

dP = Mdx +Ndy = 0

and hence implicit solutions may be obtained from

P (x, y) = C.

Example (1.2.2) Solve

(3x2 + y3ey)dx+ (3xy2ey + xy3ey + 3y2)dy = 0

HereMy = 3y2ey + y3ey = Nx.

Thus there exists P such that∂P

∂x= M,

∂P

∂y= N.

It is simplest to obtain P from

P =

∫Pxdx = x3 + xy3ey + h(y).

Differentiate with respect to y to obtain

∂P

∂y= 3xy2ey + xy3ey + h′(y) = N

and so h′(y) = 3y2 or h(y) = y3. Thus

P (x, y) = x3 + y3eyx + y3

and sox3 + y3eyx + y3 = C

provides an implicit solution.

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(d) Integrating Factors.Given

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

µ = µ(x, y) is called an integrating factor if

∂

∂y(µM) =

∂

∂x(µN)

The point of this definition is that if µ is an integrating factor, µMdx + µNdy = 0 is anexact differential equation.

Here are three suggestions for finding integrating factors:

1) Try to determine m,n so that µ = xmyn is an integrating factor.

2) IfMy −Nx

N= Q(x),

then

µ(x) = exp

[∫Q(x)dx

]is an integrating factor.

3) IfNx −My

M= R(y),

then

µ(y) = exp

[∫R(y)dy

]is an integrating factor.

(e) First Order Linear Equations.The general form of a first order linear equation is

a0(x)y′ + a1(x) = b(x)

We assume p =a1

a0

, f =b

a0

are continuous on some interval and rewrite the equation as

y′(x) + p(x)y(x) = f(x). (1.2.2)

By inspection we see thatµ = e

∫p(x)dx

is an integrating factor for the homogeneous equation. From (1.2.2) we obtain

d

dx(e

∫p(x)dxy) = f(x)e

∫p(x)dx

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and so

y(x) = e−∫p(x)dx

(∫f(x)e

∫p(x)dxdx+ c

).

Example (1.2.3) Find all solutions of

xy′ − y = x2.

If x �= 0, rewrite the equation as

y′ − 1

xy = x (1.2.3)

and so

µ = e−∫

(1/x)dx =1

|x| .

If x > 0,d

dx

(yx

)= 1

and so y = x2 + Cx. If x < 0,d

dx

(−1

xy

)= −1

and y = x2 + Cx. Hence y = x2 + Cx gives a differentiable solution valid for all x.

Now consider the initial value problem

xy′ − y = x2

y(0) = y0.

If y0 = 0, we see that there are infinitely many solutions of this problem whereas if y0 �= 0,there are no solutions. The problem in this latter case arises from the fact that when theequation is put in the form of (1.2.3) the coefficient p(x) is not continuous at x = 0. Thesignificance of this observation will be elaborated upon in the next chapter.

(e) Reduction in Order.If in the differential equation the independent variable is missing, that is, the equation is ofthe form

F (y, y′, · · · , y(n)) = 0, (1.2.4)

set

v = y′ =dy

dx,

y′′ =d

dx(y′) =

dv

dx=

dv

dy

dy

dx=

dv

dyv,

y′′′ =d

dx

(dv

dyv

)=

dv

dy

dv

dx+ v

d

dy

(dv

dy

)dy

dx

8

= v

(dv

dy

)2

+ v2d2v

dy2,

etc.

These calculations show that with this change of variables we obtain a differential equationof one less order with y regarded as the independent variable, i.e., the differential equation(1.3.1) becomes

G(y, v, v′, · · · v(n−1)) = 0.

If the dependent variable is missing, i.e.,

F (x, y′, y′′, · · · y(n) = 0,

again set v =dy

dx= y′ to obtain

G(x, v, v′′, · · · v(n−1)) = 0.

(f) Linear Constant Coefficient Equations.A homogeneous linear differential equation with constant real coefficients of order n has theform

y(n) + a1y(n−1) + · · ·+ any = 0.

We introduce the notation D =d

dxand write the above equation as

P (D)y ≡(Dn + a1D

(n−1) + · · ·+ an)y = 0.

By the fundamental theorem of algebra we can write

P (D) = (D−r1)m1 · · · (D−rk)mk (D2−2α1D+α2

1 +β21)p1 · · · (D2−2α D+α2

+β2 )p� ,

wherek∑j=1

mj + 2 ∑

j=1

pj = n.

Lemma 1.1. The general solution of (D − r)ky = 0 is

y =(c1 + c2x + · · ·+ ckx

(k−1))erx

and the general solution of (D2 − 2αD + α2 + β2)ky = 0 is

y =(c1 + c2x + · · ·+ ckx

(k−1))eαx cos(βx) +

(d1 + d2x + · · ·+ dkx

(k−1))eαx sin(βx).

Proof. Note first that (D − r)erx = D(erx)− rerx = rerx − rerx = 0 and for k > j

(D − r)(xjerx

)= D

(xjerx

)− r

(xjerx

)= jxj−1erx.

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Thus we have

(D − r)k(xjerx

)= (D − r)k−1

[(D − r)

(xjerx

) ]= j(D − r)k−1

(xj−1erx

)= · · · =

= j! (D − r)k−j (erx) .

Therefore, each function xjerx, for j = 0, 1, · · · , (k − 1), is a solution of the equation andby the fundamental theory of algebra these functions are linearly independent, i.e.,

0 =k∑j=1

cjxj−1erx = erx

k∑j=1

cjxj−1, for all x

implies c1 = c2 = · · · = ck = 0.

Note that each factor (D2 − 2αD + α2 + β2) corresponds to a pair of complex conjugateroots r = α± iβ. In the above calculations we did not assume that r is real so that for a pairof complex roots we must have solutions

e(α±iβ)x = eiβxeαx = eαx (cos(βx) + i sin(βx)) ,

and any linear combination of these functions will also be a solution. In particular the realand imaginary parts must be solutions since

1

2[eαx (cos(βx) + i sin(βx))] +

1

2[eαx (cos(βx)− i sin(βx))] = eαx cos(βx)

1

2i[eαx (cos(βx) + i sin(βx))]− 1

2i[eαx (cos(βx)− i sin(βx))] = eαx sin(βx)

Combining the above results we find that the functions

y =(c1 + c2x + · · ·+ cnx

(n−1))eαx cos(βx)

andy =

(d1 + d2x + · · ·+ dnx

(n−1))eαx sin(βx).

are solutions and, as is shown in Section 1.6, these solutions are linearly independent.

The general solution of P (D)y = 0 is given as a linear combination of the solutions for eachreal root and each pair of complex roots.

Let us consider an example which is already written in factored form[(D + 1)3(D2 + 4D + 13)

]y = 0

The term (D + 1)3 gives a part of the solution as

(c1 + c + 2x + c3x2)e−x

10

and the term (D2 + 4D + 13) corresponds to complex roots with α = −2 and β = 3 givingthe part of the solution

c4e−2x cos(3x) + c5e

−2x sin(3x).

The general solution is

y = (c1 + c + 2x + c3x2)e−x + c4e

−2x cos(3x) + c5e−2x sin(3x).

(g) Equations of Euler (and Cauchy) Type.A differential equation is of Euler type if

F (x, y′, y′′, · · · , y(n)) = F (y, xy′, · · ·xny(n)) = 0.

Set x = et so that

y =dy

dt=

dy

dx

dx

dt= y′x,

d2y

dt2− dy

dt=

d

dt(y′x)− y′x =

dy′

dx

dx

dtx + y′x− y′x = y′′x2,

etc.

In this way the differential equation can be put into the form

G(y, y, · · · , y(n)) = 0

and now the method of reduction may be applied.

An important special case of this type of equation is the so-called Cauchy’s equation ,

xny(n) + · · ·+ a1xy′ + a2y = 0,

orP (D)y ≡

(xnDn + a1x

n−1D(n−1) + · · ·+ an)y = 0.

With x = et and D ≡ dy

dtwe have

Dy =dy

dx=

dy

dt

dt

dx=

1

x

dy

dt⇒ xDy = Dy,

D2y =d

dx

(1

x

dy

dt

)=

1

x2

(d2y

dt2− dy

dt

)⇒ x2D2y = D(D− 1)y,

...

xrDry = D(D− 1)(D− 2) · · · (D− r + 1)y.

Thus we can write

P (D)y = P (D)y =

[D(D− 1) · · · (D− n + 1)

+ a1

(D(D− 1) · · · (D− n + 2)

)+ · · ·+ an−1D + an

]y = 0.

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The second order case

ax2y′′ + bxy′ + cy = 0. (1.2.5)

will arise numerous times throughout the course. We assume the coefficients are constantand x > 0 and as an alternative to the above approach we seek a solution in the form y = xr.In this way one obtains the following quadratic for the exponent r,

ar2 + (b− a)r + c = 0. (1.2.6)

There are three cases to be considered:1) If (1.2.6) has distinct real roots r1, r2 then we obtain solutions

xr1 , xr2 .

2) If r1 = α + iβ, r2 = α− iβ then solutions may be written as

xα+iβ = xαeiβ lnx = xα(cos(β lnx) + i sin(β lnx))

and similarlyxα−iβ = xα(cos(β lnx)− i sin(β lnx)).

Observe that a linear combination of solutions of (1.3.2) is again a solution and so we obtainthe solutions

1

2(xα+iβ + xα−iβ) = xα cos(β lnx)

and1

2i(xα+iβ − xα−iβ) = xα sin(β lnx).

3) If (1.2.6) has repeated roots then (b− a)2 = 4ac and

r1 = (a− b)/2a.

We seek a second solution asy = v(x)xr1

and observe that v must satisfyaxv′′ + av′ = 0.

Set w = v′ to getxw′ + w = 0

and sow = c1/x,

andv = c1 lnx + c2.

Thus in the case of repeated roots we obtain the solutions

xr1 lnx, xr1 .

One might try to verify that c1xr1 +c2xr2 , xα(c1 cos(β lnx)+c2 sin(β lnx)), and c1xr1 lnx+

c2xr1 are complete families of solutions of Euler’s equation in each of the three cases respec-

tively. That this is indeed the case will be a trivial consequence of a more general result forlinear equations that we will prove in Chapter 3.

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(h) Equations of Legendre Type.

A slightly more general class of equations than the Cauchy equations is the Legendre equa-tions which have the form

P (D)y ≡((ax + b)nDn + an−1(ax + b)n−1D(n−1) + · · ·+ a1(ax + b)D + a0

)y = 0.

These equations, just as Cauchy equations, can be reduced to linear constant coefficientequations. For these equations we use the substitution (ax + b) = et which, using the chainrule just as above, gives:

Dy =dy

dx=

dy

dt

dt

dx=

a

(ax + b)

dy

dt⇒ (ax + b)Dy = aDy,

(ax + b)2D2y = a2D(D− 1)y,

...

(ax + b)rDry = arD(D− 1)(D− 2) · · · (D− r + 1)y.

Thus we can write

P (D)y = P (D)y =

[anD(D− 1) · · · (D− n + 1)

+ a1an−1

(D(D− 1) · · · (D− n + 2)

)+ · · ·+ an−1D + an

]y = 0.

As an example consider the equation[(x + 2)2D2 − (x + 2)D + 1

]y = 0

Set (x + 2) = et; then the equation can be written as{D(D− 1)−D + 1

}y = 0

or {D2 − 2D + 1

}y = (D− 1)2y = 0.

The general solution to this problem is

y(t) = C1et + C2te

t

and we can readily change back to the x variable using t = ln(x + 2) to obtain

y = (x + 2)[C1 + C2 ln(x + 2)

].

(i) Equations of Bernoulli Type.A Bernoulli equation is a differential equation of the form

y′ + p(x)y = q(x)yn.

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I can be shown that the substitution v = y1−n changes the Bernoulli equation into the lineardifferential equation

v′(x) + (1− n)p(x)v = (1− n)q(x).

The special cases n = 0, n = 1 should be considered separately.

As an example consider the differential equation

y′ + y = ya+1

where a is a nonzero constant. This equation is separable so we can separate variables toobtain an implicit solution or we can use Bernoulli’s procedure to derive the explicit solution

y = (1 + ceax)−1/a.

The student should check that both results give this solution.

(j) Equations of Clairaut Type.

An equation in the formy = xy′ + f(y′)

for any function f is called a Clairaut equation. While, at first glance, it would appear thatsuch equation could be rather complicated, it turns out that this is not the case. As can bereadily verified, the general solution is given by

y = Cx+ f(C)

where C is a constant.

As an example the equationy = xy′ +

√4 + (y′)2

has solution y = Cx+√

4 + C2.

Sometimes one can transform an equation into Clairaut form. For example, consider theequation

y = 2xy′ + 6y2(y′)2.

If we multiply the equation by y2 we get

y3 = 3xy2y′ + 6y4(y′)2.

Now use the transformation v = y3, which implies v′ = 3y2y′, to write the equation as

v = xv′ +2

3(v′)2

whose solution is v = Cx+ 23C2 which gives y3 = Cx+ 2

3C2.

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(k) Other First Order and Higher Degree Equations.A first order differential equation may have higher degree. Often an equation is given inpolynomial form in the variable y′ and in this case we refer to the equation as having ordern if n is the highest order power of y′ that occurs in the equation. If we write p = y′ fornotational convienence, then such an equation can be written as

pn + gn−1(x, y)pn−1 + · · ·+ g1(x, y)p + g0(x, y) = 0. (1.2.7)

It may be possible to solve such equations using one of the following methods (see [?]).

Equation Solvable for p:It may happen that (1.2.7) can be factored into

(p− F1)(p− F2) · · · (p− Fn) = 0

in which case we can solve the resulting first order and first degree equations

y′ = F1(x, y), y′ = F2(x, y), · · · , y′ = Fn(x, y).

This will lead to solutions

f1(x, y, c) = 0, f2(x, y, c) = 0, · · · , fn(x, y, c) = 0

and the general solution is the product of the solutions since the factored equation can berewritten in any form (i.e., the ordering of the terms does not matter). Thus we have

f1(x, y, c)f2(x, y, c) · · · fn(x, y, c) = 0.

For example,p4 − (x + 2y + 1)p3 + (x + 2y + 2xy)p2 − 2xyp = 0

can be factored intop(p− 1)(p− x)(p− 2y) = 0

resulting in the equations

y′ = 0, y′ = 1, y′ = x, y′ = 2y.

These equations yield the solutions

y − c = 0, y − x− c = 0, 2y − x2 − c = 0, y − ce2x = 0

giving the solution

(y − c)(y − x− c)(2y − x2 − c)(y − ce2x) = 0.

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Equation Solvable for y:In this case we can write the equation G(x, y, y′) = 0 in the form y = f(x, p). In this casedifferentiate this equation with respect to x to obtain,

p =dy

dx= fx + fp

dp

dx= F (x, p,

dp

dx).

This is an equation for p which is first order and first degree. Solving this equation toobtain φ(x, p, c) = 0 we then use the original equation y = f(x, p) to try to eliminate the pdependence and obtain Φ(x, y, c) = 0.

Consider the example y = 2px + p4x2. We differentiate with respect to x to obtain

p = 2xdp

dx+ 2p + 2p4x + 4p3x2 dp

dx

which can be rewritten as (p + 2x

dp

dx

)(1− 2p3x) = 0.

An analysis of singular solutions shows that we can discard the factor (1−2p3x) and we have

(p+2xdp

dx) = 0 which implies xp2 = c. If we write the original equation as (y−p4x2) = 2px

and square both sides we have (y− p4x2)2 = 4p2x2. With this and xp2 = c we can eliminatep to obtain (y − c2)2 = 4cx.

Equation Solvable for x:In this case an equation G(x, y, y′) = 0 can be written as x = f(y, p). We proceed bydifferentiating with respect to y to obtain

1

p=

dx

dy= fy + fp

dp

dy= F (y, p,

dp

dy)

which is first order and first degree indp

dy. Solving this equation to obtain φ(x, y, c) = 0 we

then use the original equation y = f(x, p) to try to eliminate the p dependence and obtainΦ(x, y, c) = 0.

As an example consider p3 − 2xyp + 4y2 = 0 which we can write as 2x =p2

y+

4y

p.

Differentiating with respect to y gives

2

p=

2p

y

dp

dy− p2

y2+ 4

(1

p− y

p2

dp

dy

)or (

p− 2ydp

dy

)(2y2 − p3) = 0.

The term (2y2 − p3) gives rise to singular solutions and we consider only(p− 2y

dp

dy

)= 0

16

which has solution p2 = cy. We now use this relation and the original equation to eliminatep. First we have

2x− c =4y

p

which implies

(2x− c)2 = 16yy

p2=

16y

c

and finally 16y = c(2x− c)2.

1.3 Linear Nonhomogeneous Problems

Variation of parameters

Consider a nonhomogeneous second order linear equation

L(y) = y′′ + a1y′ + a2y = β

and suppose {y1, y2} is a fundamental set. Then c1y1(t)+c2y2(t) is a general solution of L(y) = 0.A method due to Lagrange, called the method of variation of parameters, for solving L(y) = β isbased on the idea of seeking a solution as

yp(t) = c1(t)y1(t) + c2(t)y2(t).

Theny′p = c1y

′1 + c2y

′2 + c′1y1 + c′2y2.

To simplify the algebra, we impose the auxiliary condition

c′1y1 + c′2y2 = 0.

Theny′′p = c1y

′′1 + c2y

′′2 + c′1y

′1 + c′2y

′2.

If we substitute into L(y) = β, we want

c1(t)(y′′1 + a1y

′1 + a2y1) + c2(t)(y

′′2 + a1y

′2 + a0y2) + c′1y

′1 + c2y

′2 = β(t).

Note that the two parenthesized expressions are zero because y1 and y2 are solutions of the homo-geneous equation. Thus we need to solve

c′1y1 + c′2y2 = 0

c′1y′1 + c′2y

′2 = β.

By Cramer’s Rule

c′1(t) =−y2(t)β1(t)

W (y1, y2)(t), c′2(t) =

y1(t)β(t)

W (y1, y2)(t).

17

Thus a particular solution is given as

yp(t) = −y1(t)

∫ t

t0

y2(s)β(s)

W (s)ds+ y2(t)

∫ t

t0

y1(s)β(s)

W (s)ds

=

∫ t

t0

[y1(s)y2(s)− y1(s)y2(s)

W (y1, y2)(s)

]β(s) ds

=

∫ t

t0

g(x, s)β(t) ds.

g(t, s) is called a Fundamental Solution.The same method works, if not as smoothly, in the general case. Consider the nth order case

L(y) = y(n) + a1y(n−1) + · · ·+ an−1y

(1) + any = β

and let {y1, . . . , yn} be a fundamental set of solutions of the homogeneous problem L(y) = 0. Asin the last section, given a basis of solutions {yj}nj=1 of the homogeneous problem Ly = 0 we seeka solution of L(y) = β in the form

yp(t) = u1(t)y1(t) + · · ·+ un(t)yn(t).

We seek a system of equations that can be solved to find u′1, . . . , u′n. To this end we note that

by applying the product rule to yp and, collecting terms carefully, we can conclude that

u′1y1 + u′2y2 + · · ·+ u′nyn = 0 ⇒ y′p = u1y′1 + u2y

′2 + · · ·+ uny

′n

u′1y′1 + u′2y

′2 + · · ·+ u′ny

′n = 0 ⇒ y′′p = u1y

′′1 + u2y

′′2 + · · ·+ uny

′′n

u′1y′′1 + u′2y

′′2 + · · ·+ u′ny

′′n = 0 ⇒ y′′′p = u1y

′′′1 + u2y

′′′2 + · · ·+ uny

′′′n

... ⇒ ...

u′1y(n−2)1 + u′2y

(n−2)2 + · · ·+ u′ny

(n−2)n = 0 ⇒ y(n−1)

p = u1y(n−1)1 + u2y

(n−1)2 + · · ·+ uny

(n−1)n

u′1y(n−1)1 + u′2y

(n−1)2 + · · ·+ u′ny

(n−1)n = β ⇒ y(n)

p = u1y(n)1 + u2y

(n)2 + · · ·+ uny

(n)n + β

which impliesL(yp) = u1L(y1) + u2L(y2) + · · ·+ unL(yn) + β = β.

Now we note that the system of equations becomes

y1 y2 · · · yn

y′1 y′2 · · · y′n

...... · · · ...

y(n−1)1 y

(n−1)2 · · · y

(n−1)n

u1

u2

...un

=

0

0

...

β

.

18

The determinant of the coefficient matrix is nonvanishing since it is the Wronskian W (t) of a setof linearly independent solutions to an n-order linear differential equation (see equation (1.5.4) inSection 1.5). Applying Kramer’s rule we can write the solutions as

u′k(t) =Wk(t)

W (t), k = 1, · · · , n

where Wk(t) is the determinant of the matrix obtained from the coefficient matrix by replacing the

kth column(i.e.,

[yk y′k · · · y

(n−1)k

]T )by the vector

[0 0 · · · β

]TIf we define

g(t, s) =n∑k=1

yk(t)Wk(s)

W (s)

then a particular solution of L(y) = β is

yp =

∫ t

t0

g(t, s) β(s) ds.

Method of Undetermined Coefficients

As we have already learned, the method of variation of parameters provides a method of rep-resenting a particular solution to a nonhomogeneous linear problem

Ly = y(n) + a1y(n−1) + · · ·+ a(n−1)y

(1) + any = f

in terms of a basis of solutions {yj}nj=1 of the linear homogeneous problem. In the special casein which the operator L has constant coefficients, we have just seen that it is possible to constructsuch a basis of solutions for the homogeneous problem. Thus given any f we can write out aformula for a particular solution in integral form

yp(t) =

∫ t

t0

g(t, s)f(s) ds.

Unfortunately, the method of variation of parameters often requires much more work than isneeded. As an example consider the problem

Ly = y′′′ + y′′ + y′ + y = 1

y(0) = 0, y′(0) = 1, y′′(0) = 0.

Example 1.1. For the homogeneous problem we have

(D3 +D2 +D + 1)y = (D + 1)(D2 + 1)y = 0

so we can takey1 = cos t, y2 = sin t, y3 = e−t.

19

Thus the wronskian is

W (t) =

∣∣∣∣∣∣∣∣∣∣

cos t sin t e−t

− sin t cos t −e−t

− cos t − sin t e−t

∣∣∣∣∣∣∣∣∣∣and we can apply Abel’s theorem to obtain

W (t) = W (0)e−∫ t0 1 ds =

∣∣∣∣∣∣∣∣∣∣

1 0 1

0 1 −1

−1 0 1

∣∣∣∣∣∣∣∣∣∣e−t = 2e−t.

Thus by the variation of parameters formula yp = u1y1 + u2y2 where

u′1 =1

2et

∣∣∣∣∣∣∣∣∣∣

0 sin t e−t

0 cos t −e−t

1 − sin t e−t

∣∣∣∣∣∣∣∣∣∣= −1

2(cos t+ sin t),

which implies

u1(t) =1

2(cos(t)− sin(t)).

Similarly, we obtain

u′2 =1

2(cos t− sin t), u3(t) =

1

2et,

which imply

u′2 =1

2(cos t+ sin t), u3(t) =

1

2et,

So we get

yp = u1y1 + u2y2

=1

2(cos t− sin t) cos t+

1

2(sin t+ cos t) sin t+

1

2ete−t

= 1

Wow! In retrospect it would have been easy to see that yp = 1 is a particular solution.Now the general solution is

y = 1 + c1 cos t+ c2 sin t+ c3e−t

and we can apply the initial conditions to determine the constants which yields

y = 1 +1

2(sin t− cos t− e−t).

20

We note that if we were to apply the method for finding a particlar solution with the properties

yp(0) = 0, y′p(0) = 0, y′′p(0) = 0,

(as given in the proof of the n order case), then we would get

yp(t) = 1− 1

2(cos t+ sin t+ e−t).

We note that the second term is part of the homogeneous solution so it can be excluded.

In any case this is a lot of work to find such a simple particular solution. In case the function fis a linear combination of:

1. polynomials,

2. polynomials times exponentials or,

3. polynomials times exponentials times sine or cosine,

i.e., if f is a solution of a linear constant coefficient homogeneous differential equation, one canapply the method of undetermined coefficients.

The method goes as follows:

1. Let L = P (D) = Dn +n∑j=1

ajDn−j

2. Let Ly = 0 have general solution yh =n∑j=1

cjyj .

3. Assume that M = Q(D) = Dm +m∑j=1

bjDm−j is a constant coefficient linear differential

operator such that Mf = 0.

4. Then L = ML is a polynomial constant coefficient differential operator.

5. Notice that if yp is any solution of Ly = f and yh is the general solution of the homogeneousproblem, then we have

L(yh + yp) = 0. (1.3.1)

6. On the other hand we can write the general solution of this problem by simply factoring Land applying the results of the previous section.

7. Note that the solution yh =n∑j=1

cjyj is part of this general solution.

21

8. So let us denote the general solution by

y =n∑j=1

cjyj +m∑j=1

djwj. (1.3.2)

9. Now we also know, by the variation of parameters formula, that there exists a particularsolution of Lyp = f .

10. But since we know the general solution of (1.3.1) is (1.3.2), this means that any particularsolution must also be a part of the full general solution of the large homogeneous problem,

i.e., yp =n∑j=1

cjyj +m∑j=1

djwj .

11. We know that Lyh = 0 so we can choose yp =m∑j=1

djwj .

12. We now substitute this expression into the original equation and solve for the constants {dj}.

Example 1.2. Example: Ly = (D2 − 2D + 2)y = t2et sin(t):

The general solution of the homogeneous equation Ly = 0 is

yh = c1et cos(t) + c2e

t sin(t)

According to the above disscussion we seek a differential operator M so that

M(t2et sin(t)) = 0.

We immediately chooseM = (D2 − 2D + 2)3

and we need to compute the general solution to the homogeneous problem

MLy = (D2 − 2D + 2)3(D2 − 2D + 2)y = (D2 − 2D + 2)4y = 0,

which implies

y = (c1 + c2t+ c3t2 + c4t

3)et cos(t) + (d1 + d2t+ d3t2 + d4t

3)et sin(t).

If we now remove the part of this function corresponding to the solution of the homogeneousproblem Ly = 0, we have

yp = (at3 + bt2 + ct)et cos(t) + (dt3 + ft2 + gt)et sin(t)

After a lengthy calculation the first derivative

y′ =((d+ a) t3 + (3 a+ b+ f) t2 + (2 b + c + g) t+ c

)et cos(t)

+((−a+ d) t3 + (−b+ f + 3 d) t2 + (−c + g + 2 f) t+ g

)et sin(t)

22

and the second derivative

y′′ =(2 dt3 + (6 d+ 6 a+ 2 f) t2

+ (6 a+ 4 b+ 4 f + 2 g) t+ 2 b+ 2 c + 2 g)et cos(t)

+(− 2 at3 + (−6 a+ 6 d− 2 b) t2

+ (−4 b + 4 f + 6 d− 2 c) t+ 2 f + 2 g − 2 c)et sin(t)

Plugging all of this into the equation yields

y′′ − 2y′ + 2y =(6 dt2 + (6 a+ 4 f) t+ 2 g + 2 b

)et cos(t) +(

−6 at2 + (−4 b+ 6 d) t− 2 c + 2 f)et sin(t)

Equating coefficients with the right hand side leads to the equations

6 d = 0

6 a+ 4 f = 0

2 g + 2 b = 0

−6 a = 1

−4 b+ 6 d = 0

−2 c + 2 f = 0

which have the solutions a = −1/6, b = 0, c = 1/4, d = 0, f = 1/4 and g = 0. Thus, a particularsolution of the equation is

y =

(−t

3

6+

t

4

)et cos(t) +

t2

4et sin(t)

The following table contains a guide for generating a particular solution when one applies themethod of undetermined coefficients. In particular, consider

Ly = f.

If f(t) = seek yp(t).

Pm(t) = c0tm + · · ·+ cm xs(a0t

m + · · ·+ am)

Pm(t)eαt ts(a0tm + · · ·+ am)eαt

Pm(t)eαt{

sin βtcos βt

tseαt[(a0t

m + · · ·+ am) cos βt

+(b0tm + · · ·+ bm) sin βt

]Example 1.3. Returning to Example 1.1 we see that the operator M = D annihilates f = 1 sowe seek a particular solution yp = 1.

23

1.4 Some Numerical Methods for ODEs

Method: Collocation Method (a Weighted Residual Method)Given a differential equation L[u] = 0 for u(ξ) with ξ ∈ Q (Q some domain) and boundary

conditions B[u] = 0, we seek an approximate solution u(ξ) = w(ξ;α) where α = {αj}Nj=1 is a setof parameters and B[w] = 0 for all choices of α.

We try to determine the α by requiring that the equation be satisfied by w at a set of pointsS = {ξj}Nj=1 ⊂ Q. Thus we arrive at a system of N equations in the N unknowns α:

L[w(ξj, α)] = 0, j = 1, · · · , N.

Example 1.4. Consider

L[u] = u′′ + u + x = 0, (1.4.1)

u(0) = 0, u(1) = 0

Suppose we choose to approximate the solution by

w(x) = α1x(1− x) + α2x(1− x2).

Note that w satisfies the boundary conditions. A straightforward calculation gives:

L[w(x)] = −α1(2− x + x2)− α2(5x + x3) + x.

Applying our collocation method at x1 = 1/3 and x2 = 2/3, we obtain the 2× 2 system

(48/27)α1 + (46/27)α2 = 1/3

(48/27)α1 + (98/27)α2 = 2/3

which has solutions α1 = 9/416 and α2 = 9/52. Thus we obtain

w(x) =9

416x(1− x) +

9

52x(1− x2).

The exact solution to this problem is

u(x) =sin(x)

sin(1)− x

and the maximum difference between y and w on [0, 1] occurs at x = .7916 with a value of .00081.

Method: Taylor SeriesThis method yields an approximate solution to a differential equation near a single point a. The

method is applicable for differential equations in which all expressions are analytic functions ofthe variables. The idea is to use the differential equation to obtain the coefficients in a Taylor seriesabout the point a. We note that this is really an application of the famous Cauchy-Kovalevskayatheorem.

24

Example 1.5. We seek an approximate solution to

y′(x) = F (x, y), y(a) = y0

in the form

y(x) =N∑j=0

y(j)(a)

j!(x− a)j.

The idea is to use the differential equation and initial condition to find the Taylor coefficients.

y(a) = y0

y′ =F (x, y) implies

y′(a) = F (a, y0)

y′′(x) =Fx(x, y) + Fy(x, y)yx implies

y′′(a) = Fx(a, y0) + Fy(a, y0)F (a, y0),

etc.

Consider

y′ = x2 − y2,

y(0) = 1

A straightforward calculation yields:

y′ = x2 − y2,

y′′ = 2x− 2yy′

y′′′ = 2− 2(y′)2 − 2yy′

y′′′′ = −6y′y′′ − 2yy′′′

from which we immediately obtain for a = 0

y′(0) = −1,

y′′(0) = 2

y′′′(0) = −4

y′′′′(0) = 20.

Thus we obtain the approximate solution

y = 1− x +2

2!x2 − 4

3!x3 +

20

4!x4

= 1− x + x2 − 2

3x3 +

5

6x4

25

Euler’s Method:This method, which can be derived many different ways, is not particularly practical but is by

far the simplest method and the starting point for many more complicated methods.Unlike the methods discussed in the earlier section, this method does not produce a function

that can be evaluated at any point but rather is gives approximate values to the solution at points inthe form {(tj, yj)}M+1

j=1 where

y(tj) ≈ yj, a = t1 < t2 < · · · < tM < tM+1 = b.

While any mesh can be used, we will select uniform mesh

tk = a+ h(k − 1), k = 1, · · · , (M + 1), h =(b− a)

M.

Assuming that y, y′ and y′′ are continuous we can apply Taylor’s theorem at t1 to obtain

y(t) = y(t1) + y′(t1)(t− t1) + y′′(c1)(t− t1)

2

2, c1 ∈ [t1, t].

Using y′(t1) = f(t1, y(t1)) and h = (t2 − t1) we get an approximation for y at t2

y(t2) = y(t1) + hf(t1, y(t1)) + y′′(c1)h2

2.

For h sufficiently small we writey2 = y1 + hf(t1, y1)

to obtain an approximate value at t2. Repeating, we have

tk+1 = tk + h, yk+1 = yk + hf(tk, yk), k = 1, · · · (M + 1) Euler’s Method.

Assuming that at each step we begin at exactly y(tk) (which won’ t be quite true), we define thelocal truncation error (LTE) to be the error obtained at a given step. In this case the LTE is

y′′(ck)h2

2.

Summing the LTE’s all the way to tM we get, roughly anyway, the so-called Global TruncationError (GTE). In this case we have

M∑k=1

y(2)(ck)h2

2≈ y(2)(c)Mh2 =

y(2)(c)

2M

(b− a)

Mh = O(h).

Thus we see that the GTE, E(h), for Euler’s method is

E(h) ≈ Ch.

From this we have

E

(h

2

)≈ C(h/2) =

1

2Ch ≈ 1

2E(h).

26

So cutting the step size in half roughly reduces the GTE by a factor of 1/2.Note that if the equation is y′ = f(t) with y(a) = 0 then this method give

y(b) ≈M∑k=1

f(tk)h

which is a Riemann sum approximation to the integral∫ b

a

f(t) dt.

Modified Euler’s Method:Once again we consider (??) and apply the fundamental theorem of calculus to get∫ t2

t1

f(t, y(t)) dt =

∫ t2

t1

y′(t) dt = y(t2)− y(t1).

This implies that

y(t2) = y(t1) +

∫ t2

t1

f(t, y(t)) dt.

Now use the trapezoid rule to evaluate the integral with step size h = t2 − t1 to obtain

y(t2) ≈ y(t1) +h

2[f(t1, y(t1)) + f(t2, y(t2))].

Unfortunately, we do not know y(t2) on the right side, so we use, for example, Euler’s method toapproximate it: y(t2) ≈ y(t1) + hf(t1, y(t1)). We thus obtain a numerical procedure by denotingy(tk) by yk for each k we set

pk = yk + hf(tk, yk), tk+1 = tk + h,

yk + 1 = yk +h

2[f(tk, yk) + f(tk+1, pk)] .

Note that if the equation were y′ = f(t) with y(0) = 0 then we would have

y(b) ≈ h

2

M∑k=1

[f(tk) + f(tk+1)]

which is exactly the trapezoid rule for numerical quadrature.In this case the LTE for the trap rule is

−y′′(ck)h3

12

and the GTE is

−M∑k=1

y(2)(ck)h3

12≈ −y

(2)(c)(b− a)

12h2 = O(h2).

27

Thus we see that the GTE E(h) for Modified Euler Method is

E(h) ≈ Ch2.

From this we have

E

(h

2

)≈ C(h/2)2 =

1

4Ch2 ≈ 1

2E(h).

So cutting the step size in half roughly reduces the GTE by a factor of 1/4. This is sometimescalled Huen’s method. See the codes a9_2.m and huen.m.

Taylor’s Method:Assuming that y has derivatives of all order we could, again by Taylor’s theorem, for any N

write

y(t+ h) = y(t) + y′(t)h+y(2)(t)h2

2+ · · ·+ y(N)(t)hN

N !(1.4.2)

with a LTE ofy(N+1)(c)hN+1

(N + 1)!.

We can adapt this result to each interval [tk, tk+1] to obtain a numerical procedure

yk+1 = yk + d1h+d2h

2

2+d3h

3

6+ · · ·+ dNh

N

N !

where dj = y(j)(tk) for j = 1, 2, · · · , N . In this case the LTE is O(hN+1) and the GTE isEN(h) ≈ ChN = O(hN). Thus we have

EN

(h

2

)=

1

2NEN(h).

Computing the values dk for a specific example is not to bad. Consider, for example, (see theexercises)

y′ =t− y

2

y(2) =2− t+ y

4

y(3) =−2 + t− y

8

y(4) =2− t+ y

16

from which we readily compute the values dj .There is a Taylor’s matlab code named taylor.m taken from John Mathews book Chapter 9. The

file a9_3.m runs an example.

Runge-Kutta Methods:

28

The dissadvantage of Taylor’s method is obvious. We have to do some analytic calculations onevery problem. To eliminate this difficulty and still have a higher order method we consider theone-step Runge-Kutta methods which are related to Taylor’s method but do not require separatecalculations.

We give a brief discussion of this method in the second order case. The fourth order case ismore common but the algebra is very messy.

In the second order case the idea is to look for a formula

yk+1 = yk + hF (tk, yk, h, f)

with F in the form

F (t, y, h, f) = γ1f(t, y) + γ2f(t+ αh, y + βhf(t, y).

where γ1, γ2, α, β are to be determined.We apply Taylor’s theorem in two variable (t, y) applied to the second term (the term containing

γ2) to obtain an approximation through the second derivative terms:

F (t, y, h, f) = γ1f(t, y)+γ2

{f(t, y)+h

[αft+βffy

]+h2

[1

2α2ftt+αβftyf+

1

2β2f 2fyy

]}+O(h3)

where

ft =∂f

∂t, fy =

∂f

∂y, etc.

Similarly, we have

y′ = f(t, y)

y′′ = ft + fyy′ = ft + fyf

y′′′ = ftt + 2ftyf + fyyf2 + ftfy + f 2

y f.

Now if we expand y(t) in a Taylor polynomial expansion about tk we have

y(tn + h) = y(tn) +y′(tn)

1h+

y′′(tn)

2h2 +

y′′′(tn)

6h3 +O(h4).

Let us denote y(k)(tn) = y(k)n , then the LTE is

LTE =y(tn+1)− y(tn)− hF (tn, y(tn), h; f)

=hy(1)n +

h2

2y(2)n +

h3

6y(3)n +O(h4)− hF (tn, y(tn), h; f)

=h[1− γ1 − γ2]f + h2[(1/2− γ2α)ft + (1/2− γ2β)fyf ]

+ h3[(1/6− 1/2γ2α2)ftt+ (1/3− γ2αβ)ftyf + (1/6− 1/2γ2β

2)fyyf2

+ 1/6fyft + f 2y f ] +O(h4)

For general f we cannot eliminate the third term but by choosing

γ1 + γ2 = 1

γ2α = 1/2

γ2β = 1/2

29

we can eliminate the first two terms. This implies that the LTE is O(h3). The system of equationsis underdetermined but we can write

γ1 = 1− γ2, α = β =1

2γ2

for arbitrary γ2.

Special Cases:For γ2 = 1/2 we get the modified Euler’s Method

yn+1 = yn +h

2[f(tn, yn) + f(tn+1, yn + hf(tn, yn))].

For γ2 = 1 we get the so-called Midpoint Method

yn+1 = yn +h

2[f(tn + h/2, yn) + f(tn+1, yn + (h/2)f(tn, yn))].

We now consider choosing γ2 in order to minimize the LTE at the O(h3) level. Note that

LTE = c(f, γ2)h3 +O(h4)

where

c(f, γ2) =

[(1

6− 1

8γ2

)ftt +

(1

3− 1

4γ2

)ftyf +

(1

6− 1

8γ2

)fyyf

2 +1

6fyft +

1

6f 2y f

].

Now applying the Cauchy-Schwartz inequality we get

|c(f, γ2)| ≤ c1(f)c2(γ2),

c1(f) =[f 2tt + f 2

tyf2 + f 2

yyf4 + f 2

y f2t + f 4

y f2

]1/2

,

c2(γ2) =

[(1

6− 1

8γ2

)2+

(1

3− 1

4γ2

)2+

(1

6− 1

8γ2

)2+

1

18

]1/2

.

We can compute the minimum of c2(·) to obtain

γ2 = 3/4, with c2(3/4) = 1/√

18.

The resulting method is usually refered to as Huen’s Method which takes the form

yk+1 = yk +h

4

[f(tk, yk) + 3f(tk +

2

3h, yn +

2

3hf(tk, yk))

].

30

There is a Huen’s matlab code named huen.m in Chapter 9 taken from John Mathews book.This same analysis can be carried out for “higher order” methods but the algebra becomes very

messy. One method that was very popular until about 1970 was the 4th order Runge-Kutta (RK4)method with a LTE of O(h5). If f does not depend on y this method reduces to simpson’ rule forquadrature. The RK4 method goes like this

f1 = f(tk, yk)

f2 = f(tk + h/2, yk + h/2f1)

f3 = f(tk + h/2, yk + h/2f2)

f4 = f(tk + h, yk + hf3)

yk+1 = yk +h

6(f1 + 2f2 + 2f3 + f4).

Assignment 3:

1. Program Example 1.4 using Maple, plot y and w and check the error given in the example.

2. Try using two different functions for example to approximate the function w in Example 1.4.

3. Try using different points x1 and x2, e.g. x1 = 1/4 and x2 = 3/4 in Example 1.4.

4. Use Maple to carry out the Taylor method up to order n for y′ = F (x, y) with y(a) = y0.Apply your program with n = 4, 6, 8 and plot the results. Use

a) F (x, y) = y2, y(1) = 1. How do your answers compare with the exact solution on[1, 2]?

b) Same question with F (x, y) = 3x2y, y(0) = 1 on [0, 1].

5. Apply Euler’s method to the following:

a) y′ = t2− y on [0, 2] with y(0) = 1. The exact answer is y = −e−t + t2− 2t+ 2. Useh = .2, .1, .05.

b) y′ =1

1− t2on [0, 1] with y(0) = 1. Note that the exact answer doesn’ t exist on this

interval. Try the methods anyway with h = .1, .05, .025.

c) y′ = e−2t − 2y on [0, 2] with y(0) = 1/10 and h = .2, .1, .05. Here y = e−2t/10 +te−2t.

6. Apply the midpoint method, the modified euler method and Huen’s method for the followingproblem with the same h in each case.

y′ = −y + t+ 1, 0 ≤ t ≤ 1, y(0) = 1.

What do you observe, i.e., how do they compare?

7. Solve the equation y′ = 1 + y2 using any method above and Taylor’s method of order 2, 4and 6 on [0, 5] with y(0) = 1 and compare your answers to y = tan(t− π/4).

31

1.5 Wronskian and Linear Independence

Consider an nth order linear nonhomogeneous equation

y(n) + a1(t)y(n−1) + a2(t)y

(n−2) + · · ·+ an(t)y = β(t) (1.5.1)

assuming ai(t), bi(t) ∈ C[a, b].Let

L(·) =dn(·)dtn

+ a1dn−1

dtn−1(·) + · · ·+ an(·).

Then (1.5.1) may be written as the linear nonhomogeneous system

L(y) = β. (1.5.2)

The operator L is said to be linear since

L(cy1(t) + c2y2(t)) = c1L(y1(t)) + c2L(y2(t)).

It easily follows that the set of solutions of the homogeneous linear system

L(y) = 0 (1.5.3)

is a vector space.The Wronskian of {yj(t)}nj=1 is defined as

W (t) ≡ W({yj(t)}nj=1

)(t) =

∣∣∣∣∣∣∣∣y1(t) · · · yn(t)

......

...y

(n−1)1 (t) · · · y

(n−1)n (t)

∣∣∣∣∣∣∣∣ . (1.5.4)

Theorem 1.1. [Abel’s Formula] If y1, . . . yn are solutions of (1.5.3) , and t0 ∈ (a, b), then

W (t) = W (t0) exp

[−

∫ t

t0

a1(s)ds

].

Thus the Wronskian of {y1, · · · yn} is never 0 or identically 0.

Proof. We compute

W ′(t) =

∣∣∣∣∣∣∣∣∣∣

y′1 · · · y′n

y′1 · · · y′n...

......

y(n−1)1 · · · y

(n−1)1

∣∣∣∣∣∣∣∣∣∣+

∣∣∣∣∣∣∣∣∣∣∣∣∣

y1 · · · yn

y′′1 · · · y′′n

y′′1 · · · y′′n...

......

y(n−1)1 · · · y

(n−1)n

∣∣∣∣∣∣∣∣∣∣∣∣∣+ · · ·+

∣∣∣∣∣∣∣∣∣∣∣∣∣

y1 · · · yn

y11

......

...y

(n−2)1 · · · y

(n−2)n

y(n)1 · · · y

(n)n

∣∣∣∣∣∣∣∣∣∣∣∣∣

32

=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

y1 · · · yn

y′1 · · · y′n...

......

y(n−2)1 · · · y

(n−2)n

−n∑j=1

αjy(n−j)1 · · · −

n∑j=1

αjy(n−j)n

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣∣∣

y1 · · · yn

y′1 · · · y′n...

......

−a1(t)y(n−1)1 · · · −a1(t)y

(n−1)n

∣∣∣∣∣∣∣∣∣∣= −a1(t)W (t).

Hence

W (t) = K exp

(−

∫ t

t0

a1(ts)ds

)or, more explicitly,

W (t) = W (t0) exp

(−

∫ t

t0

a1(s)ds

).

Definition 1.1. A collection of functions {yi(t)}ki=1 is linearly independent on (a, b) if

k∑i=1

ciyi(t) = 0 for all x ∈ (a, b)⇒ cj = 0 for j = 1, · · · , n.

Otherwise we say the set {yi(t)} is linearly dependent.

Theorem 1.2. Suppose y1, . . . , yn are solutions of (1.5.2). If the functions are linearly dependenton (a, b) then W (t) = 0 for all t ∈ (a, b). Conversely, if there is an t0 ∈ (a, b) so that W (t0) = 0,then W (t) = 0 for all t ∈ (a, b) and the yi(t) are linearly dependent on (a, b).

Proof. If the {yi(t)} are linearly dependent, then there are ci(t) not all zero such that∑i

ciyi(t) = 0 for all t ∈ (a, b)

⇒∑

ciy(k)i (t) = 0 for all t and any k.

Hence, defining

M(t) =

y1(t) · · · yn(t)

.... . .

...y

(n−1)1 (t) · · · y

(n−1)n (t)

, C =

c1...cn

,

33

the system can be written asM(t)C = 0

and since C �= 0 we see that M is singular and therefore

W (t) = det(M(t))0 for all x ∈ (a, b).

Conversely, if det(M(t)) = W (t0) = 0 then

M(t)C = 0

has a nontrivial solution. For this choice of ci’s, let

y(t) =∑

ciyi(t).

Theny(t0) = 0, y′(t0) = 0, · · · , y(n−1)(t0) = 0.

and since y is a solution of Ly = 0, from the uniqueness part of the fundamental existence uniqe-ness theorem, we must have y(t) = 0 for all x ∈ (a, b).

Example 1.6. Consider y1 = t2, y2 = t|t|. Then

W (t) =

∣∣∣∣ t2 t|t|2x 2x sgn(t)

∣∣∣∣ = 2t3 sgn(t)− 2t2|t| ≡ 0.

However, y1(t), y2(t) are not linearly dependent. For suppose

c1y1(t) + c2y2(t) = 0, for all t.

Thenc1t+ c2|t| = 0 for all t.

If t > 0,c1t+ c2t = 0⇒ c1 = −c2

while for t < 0,c1t− c2t = 0⇒ c1 = c2.

Hence c1 = c2 = 0 and so y1, y2 are linearly independent on any interval (a, b) containing 0.Thus y1, y2 are not solutions of a linear homogeneous 2nd order equation on (a, b).

1.6 Completeness of Solution for Constant Coefficient ODE

We have already learned, in Section 1.2, how to find a set of n solutions to any homogeneousequation of the form Ly = 0 with L = D(n) + a1D

(n−1) + · · · + an−1D + an. Namely, we factorthe operator into a product of factors (D− r)k and (D2− 2αD+α2 + β2)m. Having done this we

34

simply observe that the general solution of the associated homogeneous problem for each of thesetypes of operators is easy to write out. Namely, we have

(D − r)ky = 0 ⇒ y =k∑j=1

cjtj−1 ert (1.6.1)

(D2 − 2αD + α2 + β2)my = 0 ⇒ y =k∑j=1

cjtj−1 eαt cos(βt)

+k∑j=1

cjtj−1 eαt sin(βt) (1.6.2)

In the case that the coefficients ai are constant, it is possible to describe the solutions explicitlyby simply solving the homogeneous equation for each factor and adding these terms together.What we have not proved is that all such solutions give a basis for the null space of L, i.e., wehave not shown that the soutions are linearly independent. To show that these solutions are linearlyindependent is not really difficult but to do it completely rigorously and carefully is a bit lengthy.

First we note that

Lemma 1.2. If λ = α + βi is a real (β = 0 ) or complex number, then

y =

(k∑j=1

cjtj−1

)eλt

is the complete solution of (D − λ)ky = 0.

Proof. Showing that the solutions of (D− λ)ky = 0 are linearly independent amounts to showingthat (

k∑j=1

cjtj−1

)eλt = 0 for all t ∈ R⇒ cj = 0, j = 1, 2, · · · , k.

But, on noting that eλt �= 0 and dividing, this result is obvious from the fundamental theorem ofalgebra which says that a polynomial of degree k has exactly k zeros.

Lemma 1.3. If λ1 �= λ2 are two complex numbers and

p(t) =k∑j=1

cjtj−1 and q(t) =

∑j=1

djtj−1,

are two polynomials, then

p(t)eλ1t = q(t)eλ2t for all t ∈ R ⇒ p(t) = 0, q(t) = 0.

35

Proof. To see that this is true we first multiply both sides of the equation by e−λ1t so that

p(t) = q(t)e(λ2−λ1)t for all t ∈ R.

Now consider the cases in which α < 0, α > 0 and α = 0 where (λ2 − λ1) ≡ α + βi. If α < 0then (using L’Hospital’s rule in the first term)

limt→+∞

q(t)e(λ2−λ1)t = 0 while limt→+∞

p(t) = ±∞ (as ck is pos. or neg.).

So that we must have p(t) ≡ 0 and then q(t) ≡ 0. If α > 0 we repeat the same argument with thefirst limit replace by t → −∞. Finally, in the case α = 0 we divide both sides of the equation byq(t) and collect real and imaginary parts to obtain

r1(t) + ir2(t) =p(t)

q(t)= eβit = cos(βt) + i sin(βt)

where r1(t) and r1(t) are rational functions with real coefficients. Equating real and imaginaryparts we see that this would imply that

r1(t) = cos(βt), r2(t) = sin(βt)

which is impossible unless r1(t) = 0 and r2(t) = 0 since the right side has infinitely manyzeros while the left can have only a finite number. This in turn implies that p(t) = 0 and alsoq(t) = 0.

Lemma 1.4. If < > 0, λ1 �= λ2 are real or complex numbers and

(D − λ2) (p(t)eλ1t

)= 0

where p(t) is a polynomial, then p(t) ≡ 0.

Proof. We know that every solution of (D − λ2) y = 0 can be written as y =

(q(t)eλ2t

)for

some polynomial q(t) of degree at most (< − 1). So the equestion is whether or not there exists apolynomial q(t) so that

p(t)eλ1t = q(t)eλ2t.

We note that this is only possible when p(t) = 0 and q(t) = 0 by Lemma 1.3.

Lemma 1.5. If p(t) is any polynomial of degree less than or equal (n− 1) then

(D − λ1)m

(p(t)eλ2t

)= q(t)eλ2t

where q(t) is a polynomial of degree at most the degree of p(t).

Proof. Consider the case m = 1. We have

(D − λ1)(p(t)eλ2t

)= q(t)eλ2t

where q(t) = p′(t) + (λ2 − λ1)p(t) which is a polynomial of degree p(t). You can now iterate thisresult for general < > 0.

36

Lemma 1.6. If L(y) = y(n)+a1y(n−1)+ · · ·+any has real coefficients and p(r) = rn+a1r

(n−1)+· · ·+ an. Then p(z) = p(z) for all z ∈ C. Therefore if p(α + βi) = 0 then p(α− βi) = 0.

Proof. For every z1, z2 ∈ C, we have z1 + z2 = z1 + z2 and z1z2 = z1z2 which also implieszn1 = z1

n.

From Lemma 1.6 we know that for a differential operator L with real coefficients, all complexroots must occur in complex conjugate pairs (counting multiplicity) and from Lemma 1.2 we knowthat for a pair of complex roots λ = α+ βi each of multiplicity k, a set of 2k linearly independentsolutions is given for j = 0, · · · , (k − 1) by

tjeλt = tjeαt(

cos(βt) + i sin(βt)

).

tjeλt = tjeαt(

cos(βt)− i sin(βt)

).

From this we see that there is a set of real solutions given as a linear combination of these solutionsby

tjeαt cos(βt) =1

2tj

(eλt + eλt

),

and

xjeαt sin(βt) =1

2itj

(eλt − eλt

).

We already know from Lemma 1.2 that tjeλt and tjeλt are linearly independent. Suppose wehave a linear combination

cjtjeαt cos(βt) + djt

jeαt sin(βt) = 0.

This would imply that(cj − dji)

2tjeλt +

(cj + dji)

2tjeλt = 0,

but since these functions are independent this implies

(cj − dji) = 0, (cj + dji) = 0, which implies cj = dj = 0.

Combining these results we have the main theorem:

Theorem 1.3. If L = y(n) + a1y(n−1) + · · · + any has real coefficients and we assume that the

polynomial p(r) = rn + a1r(n−1) + · · ·+ an has zeros given by

r1, r1, r2, r2, · · · , r , r , r2 +1, · · · , rswhere rj = αj +βji, j = 1, · · · , <, αj, βj ∈ R, βj �= 0 and rj for j = 2<+1, · · · , s are real. Letrj have multiplicity mj for all j. Then if pj(t) and qj(t) denote arbitrary polynomials (with realcoefficients) of degree (mj − 1), the general solution of Ly = 0 can be written as

y = ∑

j=1

eαjt [pj(t) cos(βjt) + qj(t) sin(βjt)] +s∑

j=2 +1

pj(t)erjt.

37

Proof. We need only prove that all the functions making up this general linear combination arelinearly independent. We already know that each particular term, i.e., a term of the form pj(t)e

rjt

or eαjt [pj(t) cos(βjt) + qj(t) sin(βjt)] consists of linearly independent functions. Note also thatby rewriting this last expression in terms of complex exponentials, we have the functions pj(t)e

rjt

and pj(t)erjt. Thus let us suppose that we have a general linear combination of the form

m∑j=1

pj(t)erjt = 0, for some m,

where all we assume is that ri �= rj for i �= j. We want to show this implies that every polyniomialpj ≡ 0. We prove this by induction:

1. The case s = 1 have already done.

2. Assume that the statement holds for s = k − 1, i.e.,k−1∑j=1

pj(t)erjt = 0 implies that every

pj(t) ≡ 0.

3. Assume thatk∑j=1

(pj(t)e

rjt

)= 0. We now apply (D − rk)

mk to this expression and note

that (D − rk)mk

(pk(t)e

rkt)

= 0 so that the sum reduces to

k−1∑j=1

(D − rk)mk

(pj(t)e

rjt

)= 0.

By Lemma 1.5 this sum can be written as

k−1∑j=1

qj(t)erjt = 0

where(D − rk)

mk(pj(t)e

rjt)

= qj(t)erjt

By the induction hypothesis we have qj(t) = 0 for all j = 1, · · · , (k − 1). But this impliesthat

(D − rk)mk

(pj(t)e

rjt)

= 0, j = 1, · · · , (k − 1)

which by Lemma 1.4 implies that pj(t) = 0 for all j = 1, · · · , (k − 1). Finally we see thatthe original expression reduces to (

pk(t)erkt

)= 0

which implies that pk(t) = 0.

38

Assignment 1

1. Solve the differential equations.

(a) y′ + 2xy + xy4 = 0

(b) y′ =y

x+ sin

y − x

x

(c) (2x3y2 + 4x2y + 2xy2 + xy4 + 2y)dx+ 2(y3 + x2y + x)dy = 0

(d) (y − y′x)2 = 1 + (y′)2

(e) x2yy′′ + x2(y′)2 − 5xyy′ + 4y2 = 0

(f) y′ = (−5x + y)2 − 4

(g) xydx + (x2 + y2)dy = 0

(h) y =y′x

2− 8

(x

y′

)2

(i) x =y

3y′− 2y′y2

(j) (y′)4 − (x + 2y + 1)(y′)3 + (x + 2y + 2xy)(y′)2 − 2xyy′ = 0

2. Solve 1 + yy′′ + (y′)2 = 0.

3. Consider a differential equation M(x, y) dx + N(x, y) dy = 0 and assume that there is aninteger n so that M(λx, λy) = λnM(x, y), N(λx, λy) = λnN(x, y) (i.e., the equation ishomogeneous).

Then show that µ = (xM + yN)−1 is an integrating factor provided that (xM + yN) is notidentically zero. Also, investigate the case in which (xM + yN) ≡ 0.

4. Solve the equations

(a) (x4 + y4) dx− xy3 dy = 0

(b) y2 dx+ (x2 − xy − y2) dy = 0

5. Find the general solution for the differential equation with independent variable x.

(a) y′′′ + 2y′′ − 8y′ = 0

(b) y′′′ + 3y′′ + 28y′ + 26y = 0

(c) y′′′ − 9y′′ + 27y′ − 27y = 0

(d) y(4) + 4y′′ + 4y = 0

(e) (D − 1)2(D + 3)(D2 + 2D + 5)2y = 0

6. Solve the initial value problem

y′′′ − y′′ − 4y′ + 4y = 0, y(0) = −4, y′(0) = −1, y′′(0) = −19.

39

7. Find the general solution

(a) y′′ + 2y′ + 10y = 25x2 + 3.

(b) y′′ + y′ − 6y = −6x3 + 3x2 + 6x.

(c) y′′ + 10y′ + 25y = e−5x.

(d) y′′ + 1.5y′ − y = 12x2 + 6x3 − x4 with y(0) = 4 and y′(0) = −8.

(e) y′′ + 2y′ + y = e−x cos(x).

(f) y′′ − 2y′ + y = ex/x3.

(g) (x2D2 − 2xD + 2)y = x3 cos(x).

(h) (x2D2 + xD − 9)y = 48x5.

8. Find the general solution.

a) y′′ + y = sinx sin 2x

b) y′′ − 4y′ + 4y = ex + e2x

c) y′′ + 9y = sec 3x

d) y′′ − 2y′ + y =ex

x3

40