Prelims Introductory Calculus 2012MT

7/30/2019 Prelims Introductory Calculus 2012MT

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Prelims: Introductory Calculus I (ODE part)

2012 Michaelmas Term. Lecturer: Z. Qian

Mathematical Institute, Oxford

October 18, 2012

1 Standard integrals, integration by parts

It is important to grasp some basic techniques for evaluating integrals such as the method ofsubstitutions, integration by parts etc. You should refer Richard Earls lecture notes, posted onthe course web page, about a few standard substitutions.

Before we give some examples, let us introduce some notations. Recall that in A-level math,we write integral of a function f as

f(x)dx (indefinite integral),

10 f(x)dx etc. f(x) is called

the integrand, and the expression f(x)dx is called a differential form (of first order). It will bebeneficial to introduce the notion of differentials. If y = f(x) is a function of one variable x onsome interval, then dy = df(x) f(x)dx is called the differential of f. The fundamental theoremin calculus says that

df(x) = f(x)dx = f(x) + C

where C is an arbitrary constant.The chain rule for derivatives implies that the differential of a function is invariant under

substitutions. More precisely, suppose y = f(x) is a function of x, making substitution x = g(t) sothat y = f(g(t)) is a function of t . Then dx = g(t)dt so that

dy = f(x)dx = f(g(t))g(t)dt

=d

dtf(g(t))dt. [Chain rule]

That is df(x) = df(g(t)) if x = g(t), in other words, when we work out the differential df(x) itdoesnt matter if we consider x as a variable or as a function of another variable. This principle

also applies to differential forms of first order. The substitution method then can be summarizedas the following equality

f(x)dx = [Substitute x = (t)]

f((t))d(t)

=

f((t))(t)dt.

There is a similar version for definite integrals.

Example 1.1 Evaluate I =

10

dx42xx2 .

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The integral is close to the integral

dx1x2 which equals sin

1 x up to a constant, so we attemptto use this known integral. By completing square we may write

4 2x x2 = 5 (x + 1)2

= 5

1

x + 15

2,

making substitution t = x+15

, dx =

5dt, where t : 15 2

5, we have

I =1

5

25

15

5dt

1 t2 = 2

5

15

dt1 t2

= sin12

5 sin1 1

5.

Now let us recall the technique of integration by parts, which is in many aspects the soul ofthe analysis. Integration by parts is the integral form of the product rule for derivatives, since(f g) = fg + f g so that

f(x)g(x) =

g(x)f(x)dx +

f(x)g(x)dx

rearranging the terms to obtainf(x)g(x)dx = f(x)g(x)

g(x)f(x)dx.

Similarly we have ba

f(x)g(x)dx = f(x)g(x)|ba ba

g(x)f(x)dx,

or in terms of differentials we can rewrite the preceding formula asba

f(x)dg(x) = f(x)g(x)|ba ba

g(x)df(x).

However there is no general rule to tell us how to split an integrand into g(x)f(x).

Example 1.2 Consider I =

xexdx. Then

I =

xdex = xex exdx= (x 1)ex + C.

Example 1.3 Now let us consider In =

xnexdx where n = 1, 2, 3, . Using integration by parts

In =

xndex = xnex

exdxn

= xnex n

xn1exdx

= xnex nIn1

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which gives a reduction formula. By repeating the use of integration by parts, one can eventuallywork out the result. For example

I2 = x2ex 2I1 =

x2 2(x 1) ex + Cand

I3 = x3ex 3I2

=

x3 3 x2 2(x 1) ex + Cetc.

Example 1.4 Consider In =

cosn xdx where n is a non-negative integer. Split the integrandcosn x into cosn1 x cos x = cosn1 x(sin x), and perform integration by parts. Then

In =

cosn1 x(sin x)dx =

cosn1 xd sin x

= cosn1 x sin x

sin xd cosn1 x

= cosn1 x sin x (n 1)

sin x cosn2 x( sin x)dx

= cosn1 x sin x + (n 1)

sin2 x cosn2 xdx.

Applying the identity sin2 x = 1 cos2 x in the last integral, we obtain

In = cosn1 x sin x + (n 1)

(1 cos2 x)cosn2 xdx= cosn1 x sin x + (n 1)In2 (n 1)In.

Collecting In together to obtain

nIn = (n 1)In2 + cosn1 x sin x

so that

In =n 1

nIn2 +

1

ncosn1 x sin x

which reduces the calculation of In to I0 or I1, both are easy to evaluate. For example/20

cosn xdx =n 1

n

/20

cosn2 xdx +1

ncosn1 x sin x

/20

=n 1

n

/20

cosn2 xdx =

=

n1n

n3n2 /20 cos xdx if n = old,

n1n

n3n2 /20 dx if n = even.

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Example 1.5 Consider I =

ex sin xdx. We have

I =exd cos x = ex cos x + ex cos xdx

= ex cos x +

exd sin x

= ex cos x + ex sin x

ex sin xdx

= ex cos x + ex sin x Iso that

2I = ex cos x + ex sin x + C.

2 First order differential equations

A (ordinary) differential equation is an equation involving an independent variable x, a function

y(x) and its derivatives:F(x,y,y, , y(n)) = 0.

By solving the highest order derivative y(n) in terms of lower order derivatives y(k) for k < n andx, the above equation may be written as

y(n) = f(x,y,y, , y(n1)). (2.1)Such an equation is called an n-th order differential equation. Ifn = 1, then it is called a first orderdifferential equation. Thus a first order differential equation has the general form y = f(x, y), orimplicitly F(x,y,y) = 0.

A function y = (x) defined on some interval J is called a solution of (2.1) if

(n)(x) = f(x, (x), (x), , (n1)(x)) x J.A function y = (x) which contains n independent arbitrary constants C1, , Cn is called thegeneral solution of (2.1) if 1) it is a solution for any arbitrary choice of C1, , Cn, 2) any solutionof (2.1) has this form.

The concept of general solutions is not very useful. We are often interested in the so-called initialproblems or boundary problems. Observe that in order to determine the constants C1, , Cn ingeneral we need n conditions which appear as initial conditions. More precisely, an initial conditionfor n-th order differential equation (2.1) may be formulated as

y(x0) = y0, , y(n1)(x0) = yn1where x0 J and y0, , yn1 are given data.

A differential equation is called an (inhomogenous) linear differential equation, if it is linear iny, y, , y(n), so that a linear differential equation can be written as the following general form

an(x)y(n) + an1(x)y

(n1) + + a0(x)y = h(x)where an, , a0 and h are functions of x. If h 0, then the linear equation is homogenous.

A first order linear differential equation can be thus put in the following general form

y + p(x)y = q(x).

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2.1 Separable first order DE

Consider a first order differential equation dydx = f(x, y). It is separable iff(x, y) = a(x)b(y), so that

dydx = a(x)b(y). Dividing the equation by b(y) and multiplying it by dx to separate the variables xand y and write the equation to be

dy

b(y)= a(x)dx.

Integrating both sides of the equation to obtain the solution given bydy

b(y)=

a(x)dx

which gives in general solutions of a separable equation implicitly. If y0 is a root to b(y) = 0, thenclearly the constant function y = y0 is also a solution.

Example 2.1 Find the general solution to

x(y2 1) + y(x2 1) dydx

= 0.

The equation is separable and can be rearranged as

xdx

x2 1 +ydy

y2 1 = 0.

After integration we obtainln |x2 1| + ln |y2 1| = C

(C is a constant), which can be put in the form

(x2 1)(y2 1) = C.The constant functions y = 1 or y = 1 are solutions but are already included in the above general

form with C = 0.

Example 2.2 Find the solution to (1 + ex)yy = ex satisfying the initial condition that y(0) = 1.The equation is separable:

ydy =ex

1 + exdx.

After integration we obtain the general solution

1

2y2 = ln(1 + ex) + C.

To match the initial condition, we set x = 0 and y = 1 in the general solution to determine theconstant C = 12 ln 2, so that 12y2 = ln(1 + ex) + 12 ln 2. After simplification we have

y2 = lne

4(1 + ex)2

.

Some differential equations of first order can be transformed by proper substitutions to separableequations.

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Example 2.3 Find the general solution to y = sin(x + y + 1).

Let u(x) = x + y(x)+1 so that u = 1 + y. The original equation can be formulated a DE of u,namely u = 1 + sin u which is separable. Dividing the equation by 1 + sin u and write the equationas

du

1 + sin u= dx.

Integrating the equation we obtain du

1 + sin u=

dx.

Let us evaluate the integral on the left hand side.

du1 + sin u

= (1 sin u)du(1 + sin u)(1 sin u)

=

(1 sin u)du

1 sin2 u =

(1 sin u)ducos2 u

=

du

cos2 u

sin udu

cos2 u

=

du

cos2 u+

d cos u

cos2 u= tan u 1

cos u+ C.

Therefore

tan u 1cos u

= x + C

or in terms of y and x, the solution is given by

tan(x + y + 1) 1cos(x + y + 1)

= x + C

orsin(x + y + 1) 1 = (x + C)cos(x + y + 1).

We also have solutions x + y + 1 = 2n 2 where n =integers.

2.2 Homogenous equations

Consider a first order differential equation dydx = f(x, y). If the function f(x, y) (of two variables)is homogenous, i.e. f(x, y) = h( y

x

) where h is a function of one variable, then we can make a

substitution u(x) = y(x)x so that y = xu. The product rule gives thatdydx = u + x

dudx , and the

equation may be written as

u + xdu

dx= f(u)

which is separable.

Example 2.4 Find general solutions to xy =

x2 y2 + y. The equation, by dividing x bothsides, is homogenous

y =

1 y

x

2+

y

x

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so we make substitution u = yx and change equation to be

u + x

du

dx =

1 u2

+ u.

Rearrange the equation: du1u2 =

dxx . Integrating both sides to obtain

sin1 u = ln |x| + Cor in terms of y, general solutions are given by sin1( yx) = ln |x|+ C, together with solutions yx = 1and yx = 1.

Some differential equations of first order can be transformed into homogenous ones by simplesubstitutions.

For example, consider the following type of first order differential equations

dy

dx= f

a1x + b1y + c1a2x + b2y + c2

.

If c1 = c2 = 0 then the equation is homogenous, so we consider the case that c1 or c2 does notvanish. If a1 b1a2 b2

= 0and b1 = 0, then we make substitution u(x) = a1x+b1y(x) to transform the equation to a separableone. For the the case where

a1 b1a2 b2 = 0

we make translation x = t + k and y = z + l such thata1k + b1l + c1 = 0,a2k + b2l + c2 = 0.

Consider t as a new independent variable, and z as a function of t, then

z(t) = y(x) l = y(t + k) ltherefore, by chain rule,

dz

dt=

dy

dx.

The differential equation we are interested becomes

dz

dt= f

a1t + b1z

a2t + b2z

which is homogenous.

Example 2.5 Find the general solution to

y = 2

y + 2

x + y 12

.

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Solve the linear system

l + 2 = 0,k

+l 1 = 0

to obtain l = 2 and k = 3. Let x = t + 3 and y = z 2. Then the differential equation can bewritten as

z = 2

z

t + z

2

which is homogenous. Now making standard substitution u(t) = z(t)t , so that z = u + tu and

u + tdu

dt=

2u2

(1 + u)2

which is separable. Rearrange the equation

tdu

dt=

2u2 u(1 + u)2(1 + u)2

= u(1 + u2)

(1 + u)2

and separate the variables to obtain

(1 + u)2

u(1 + u2)du = dt

t. (2.2)

Since

(1 + u)2

u(1 + u2

)

du = 1u

+2

1 + u2 du

= ln |u| + 2 tan1 u

therefore, by integrating the equation (2.2) we obtain

ln |u| + 2 tan1 u = ln |t| + C.

In terms of x and y the general solution is given by

ln |y + 2| + 2 tan1 y + 2x 3 = C.

2.3 Linear differential equations of first order

Consider a linear differential equation of first order

dy

dx+ p(x)y = q(x) (2.3)

where p and qare two continuous functions. The corresponding homogenous equation dzdx+p(x)z = 0is separable, and has the general solution

z(x) = Cep(x)dx

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where

p(x)dx is a primitive of p(x), and C is an arbitrary constant. It follows that z(x)ep(x)dx

is a constant, so thatd

dx

z(x)ep(x)dx

= 0

which is in turn equivalent to the homogenous equation z + p(x)z = 0.Next we consider the inhomogenous equation (2.3). The previous discussion suggests to consider

the differential of y(x)ep(x)dx, and by employing the product rule for derivatives, we obtain

d

dx

y(x)e

p(x)dx

= ep(x)dx

dy

dx+ p(x)y

= q(x)e

p(x)dx (2.4)

so by integrating the equation both sides we obtain

yep(x)dx =

q(x)ep(x)dxdx + C

dividing by ep(x)dx the equality to obtain the general solution of (2.3)

y = ep(x)dx

q(x)e

p(x)dxdx + C

. (2.5)

The function ep(x)dx which is multiplied to y to form ye

p(x)dx is called an integrating factor

to the inhomogenous equation (2.3).We may describe the above procedure to obtain general solutions for first order linear differential

equations as following, which includes an idea that can be applied to other different situations, thusare worthy of learning.

Observe that z(x) = ep(x)dx is a non-trivial solution to the corresponding homogenous equa-

tion z +p(x)z = 0, in order to obtain the general solution to the inhomogenous one (2.3), we makeuse of the solution z(x): making substitution

u(x) =y(x)

z(x)(2.6)

(which is a standard substitution as long as z is a known function which has some thing to do withthe differential equation we are interested. We will use this idea in several occasions later on), andturn (2.3) into a differential equation in u. Of course, according to the explicit form of z(x) we

have u(x) = y(x)ep(x)dx

and (2.4) just says that

u = q(x)ep(x)dx

which can be integrated to obtain the solution u.

Example 2.6 Solve differential equation y + 2xy = 2xex2

.

First work out an integrating factor r(x) = e2xdx = ex

2

. Multiplying r(x) both sides theequation we obtain

ex2

y + 2xex2

y = 2x

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that isd

dxex

2

y = 2x.

After integration we haveex

2

y = x2 + C

so that y =

x2 + C

ex2

is the general solution.

Example 2.7 Bernoullis equation is a non-linear first order equation

dy

dx+ p(x)y = q(x)yn

where n = 0 or 1 (but not necessary an integer).Dividing by yn, the equation becomes

1

yndy

dx+ p(x)y1n = q(x)

By using transformation z = y1n the equation is transformed to a linear equation

dz

dx+ (1 n)p(x)z = (1 n)q(x)

so that

y1n = e(1n)p(x)dx

(1 n)

q(x)e(1n)p(x)dxdx + C

.

3 Linear differential equations

Differential equations of second order play a special role in science. Many physical equations aresecond order ordinary or partial differential equations, such as the dynamics described by Newtonslaw of gravity, fluid dynamics which are determined by the fluid equations: Navier-Stokes equations.

3.1 Structure of general solutions to linear differential equations

Let us first describe the structures of solutions to linear differential equations. Recall the generallinear differential equation of order n is an equation that can be written

an(x)y(n) +

+ a1(x)y

+ a0(x)y = f(x) (3.1)

where ai are continuous functions (on some interval) and an = 0.Suppose yp is a particular solution of (3.1), then clearly, y is a solution to (3.1) if and only if

y yp is a solution to the corresponding homogenous linear DE of n-th order

an(x)y(n) + + a1(x)y + a0(x)y = 0. (3.2)

If y1 and y2 are two solutions to (3.2), then so is y1 + y2, and moreover, there are n solutionsy1, , yn of (3.2) such that the general solution

y = C1y1 + + Cnyn

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where C1, , Cn are arbitrary constants. That is, the collection of all solutions to a homogenouslinear equation of n-th order is a vector space of dimension n. It follows that the general solution

to (3.1) is given by y = C1y1 + + Cnyn + y0where yp is a particular solution of (3.1), C1y1 + + Cnyn is the general solution to the corre-sponding homogenous equation (3.2).

Let us investigate again the general observation we have used to solve general linear differentialequation of first order. That is, if there is a non-trivial function z(x) which has some connectionto the differential equation we are interested (for example, for a linear equation, the function maybe a solution to the corresponding homogenous equation), we can make use of the known function

in a canonical way by making substitution that u(x) = y(x)z(x) .Obviously the constant zero function is a trivial solution to any homogenous linear equation

which of course give us no additional information about the solutions. Suppose however we know,

say by inspection, a non-trivial solution z(x) to the homogenous equation (3.2), then we may reducethe equation to a lower order differential equation. Let us demonstrate this idea for homogeneoussecond order differential equations, for simplicity.

Suppose z(x) = 0 is a non-trivial solution to a homogenous linear differential equation of secondorder

p(x)d2y

dx2+ q(x)

dy

dx+ r(x)y = 0. (3.3)

Making the standard substitution u(x) = y(x)z(x) , so that, y = uz. Then y = uz + uz and

y = uz + 2uz + uz, substitute these equations to (3.3) we obtain

p(x) uz + 2uz + uz+ q(x) uz + uz+ r(x)uz = 0.Rearrange the above equation and use the fact that z is a solution to (3.3)

p(x)z(x)d2u

dx2+

2p(x)z(x) + q(x)z(x) du

dx= 0 (3.4)

which is a homogenous differential equation of first order for unknown function dudx .

Example 3.1 Verify that z(x) = 1x is a solution to

xy + 2(1 x)y 2y = 0hence find its general solution.

Since z = x2 and z = 2x3 we can easily see that z is a solution. Making substitutiony(x) = 1xu(x) in the equation we obtain a differential equation for u:

x1

x

d2u

dx2+

2x2x + 2(1 x) 1

x

du

dx= 0.

Let w = dudx and simplify the above equation:

dw

dx 2w = 0

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which is separable, and has a general solution w(x) = C1e2x. Integrating w to obtain

u(x) =

w(x)dx = C1e2x

+ C2

so that

y(x) =1

x

C1e

2x + C2

(3.5)

is the general solution, where C1 and C2 are arbitrary constants.

Example 3.2 Find the general solution to the inhomogenous linear equation

xy + 2(1 x)y 2y = 12x.

We have found the general solution to the corresponding homogenous equation which is givenby (3.5), thus, according to the structure of solutions to linear equations, we only need to find aparticular solution. Since the coefficients of the equation are all polynomials in x so we may lookfor a solution with a form y(x) = ax+b where a, b are constants. Plugging into the equation y = 0,y = a and y = ax + b into the equation

2a(1 x) 2(ax + b) = 12x

so we should have 2a2b = 0 and 2a2a = 12 so that a = 3 and b = 3. Thus y0(x) = 3x3is a particular solution, and the general solution thus is given by

y(x) =1

xC1e2x + C2

3x

3.

3.2 Linear ODE with constant coefficients

For homogenous linear ODE with constant coefficients:

y(n) + an1y(n1) + + a1y + a0y = 0 (3.6)

where an1, , a0 are constants, we can construct its general solution if we can find the roots tothe auxiliary equation

mn + an1mn1 + + a1m + a0 = 0. (3.7)

The auxiliary equation comes from the following observation. Since the derivative of emx is memx

it is thus reasonable to search for a solution y = emx. Substitute y(k) = mkemx into (3.6) we havemn + an1m

n1 + + a1m + a0

emx = 0

thus emx is a solution if and only if m is a root to (3.7) and as long as m is real. Ifm = + i is acomplex root of the auxiliary equation, then since the coefficients an1, , a0 are real numbers, sothat m = i is also a root. Now the complex functions emx and emx both satisfy the differentialequation (3.6) so that the real part and imaginary parts of

emx = ex cos(x) + iex sin(x)

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(Eulers equation) are solutions of (3.6), i.e. if m = + i is a complex root of the auxiliaryequation, then

y1(x) = e

x

cos(x)and

y2(x) = ex sin(x)

are a pair of solutions of (3.6).Ifm is a repeated root to the auxiliary equation with multiplicity k 2, then emx, xemx, , xk1emx

are solutions. The similar conclusion is valid for complex roots. We therefore are able to constructn linearly independent solutions to (3.6) via the roots to the auxiliary equation.

Example 3.3 Consider the harmonic motion described by

d2y

dx2 + 2

y = 0

where = 0 is real. The auxiliary equation is m2 + 2 = 0 which has two complex roots m = iand m. So we have two independent solutions cos x and sin x and the general solution

y(x) = A cos x + B sin x

where A, B are arbitrary constants.

Example 3.4 Solve the equation

d3y

dx3 4d2y

dx2 +

dy

dx + 6y = 0.

The auxiliary equationm3 4m + m + 6 = 0

has roots1, 2, 3 so the general solution

y(x) = C1ex + C2e

2x + C3e3x.

The situation for second ED with constant coefficients is particularly simple. Consider thehomogenous linear equation

d2y

dx2

+ ady

dx

+ by = 0 (3.8)

where a, b are two real numbers.

Theorem 3.5 Suppose the auxiliary equation

m2 + am + b = 0

has two roots m1 and m2.1) If m1 = m2 are real, then the general solution is given by

y(x) = C1em1x + C2e

m2x .

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2) If m = m1 = m2 is a repeated real root, then the general solution

y(x) = (C1 + C2x) emx.

3) If m1 = + i is a complex root (= 0) so that m2 = i, then the general solution

y(x) = ex (C1 cos x + C2 sin x) .

Proof. Note that a = (m1 + m2) and b = m1m2. We consider 1) and 2) first. In this caseem1x is a solution, so we make substitution y(x) = u(x)em1x in the differential equation. Since

y =

u + m1u

em1x

and

y =

u + 2m1u + m2

1u

e

m1x

we obtain u + 2m1u

+ m21u

+ a

u + m1u

+ bu = 0.

Using the fact that 1 is a root and that a = (m1 + m2), we have

u (m2 m1) u = 0.

Thus, if m2 m1 = 0,u(x) = C1e

(m2m1)x

and integrating the equation to obtain

u(x) = C1e(m2m1)x + C2

which proves 1). If m2 m1 = 0 then u = 0 so by integrating twice to obtain

u(x) = C1 + C2x

which shows 2).

Example 3.6 Solve the differential equation

d2y

dx2

2

dy

dx

+ 5y = 0.

The auxiliary equation m22m +5 = 0 has complex root m = 1+ 2i and m, so the general solution

y(x) = C1ex cos2x + C2e

x sin2x.

Next we give some examples for inhomogenous linear equations.

Example 3.7 Solve the equationd2y

dx2+ 4y = sin3x.

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It is easy to find the general solution to the corresponding homogeneous equation

d2y

dx2 + 4y = 0

whose auxiliary equation m2 + 4 = 0 has two complex roots 2i. Since sin3x is the imaginarypart ofe3ix, and 3i is not a root of the auxiliary equation. Thus we search for a particular solutionyp(x) = A sin3x. Plugging into the equation we find A = 15 . Hence the general solution

y(x) = C1 cos2x + C2 sin2x 15

sin3x.

Example 3.8 Considerd2y

dx2+ 4

dy

dx+ 4y = sin 3x.

The auxiliary equation m2 + 4m + 4 has 2 and 2. There is no particular solution with a formA sin3x by a simple inspection, instead we look for a particular solution

yp(x) = A cos3x + B sin3x.

Then9A + 12B + 4A = 0

and9B 12A + 4B = 1.

Thus A = 12169

, B = 5169

.

The general solution

y(x) = C1e2x + C2e

2x 12169

cos3x 5169

sin3x.

Example 3.9 Let us now consider

d2y

dx2+ 4y = sin2x.

We have seen that cos 2x is a solution to the corresponding homogenous equation, so we lookfor a particular solution

yp(x) = Ax cos2x + Bx sin2x.

Then B = 0 and A = 14 , so the general solution

y(x) = C1 cos2x + C2 sin2x 14

x cos2x.

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Example 3.10 Find a particular solution to

d2y

dx2 + 4y = sin x + sin 2x .

By a simple inspection, y1 =13 sin x is a particular solution to

d2y

dx2+ 4y = sin x

and we know from the previous example y2 = 14x cos2x is a particular solution tod2y

dx2+ 4y = sin2x.

Thus

yp = 13

sin x 14

x cos2x

is a particular solution.

Example 3.11 Let us consider inhomogenous linear equation

d2y

dx2 3 dy

dx+ 2y = f(x)

where f(x) is a given function. The auxiliary equation m2 3m + 2 = 0 has two real roots 1 and2, so the general solution to the corresponding homogenous equation is C1e

x + C2e2x.

1) Suppose f(x) = sin x which is the imaginary part of eix, since i is not a root of the auxiliary

equation, so we may search for a particular solution yp = A sin x + B cos x, but not just A sin xwhich is not good.

2) f(x) = e3x. Since 3 is not a root of the auxiliary equation, so search for a particular solutionyp = Ae

3x. If however f(x) = e2x, then we may attempt a particular solution yp = Axe2x as e2x is

a solution to the corresponding homogenous equation.3) f(x) = xe2x. Since 2 is a root of the auxiliary equation, so we may search for a particular

solution in a form yp = (Ax2 + Bx)e2x (we have included Bxe2x as well, since e2x is a solution to

the homogenous equation, but not xe2x).4) f(x) = ex sin x which is the imaginary part of e(1+i)x and 1 + i is not a root of the auxiliary

equation, so we may search for a particular solution yp = (A cos x + B sin x)ex.

5) f(x) = sin2 x. Since sin2 x = 12cos2x, so we may attempt a particular solution with a formyp = A + B cos2x + Csin2x.

4 Some facts about matrices

An m n matrix A is an array of numbers arranged into m rows and n columns

A =

a11 a12 a1na21 a22 a2n

...... ...

am1 am2 amn

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and simply written as A = (aij), where aij is the entry in the ith row and jth column. If m = n,then A is called a square matrix.

Let us concentrate on 22 matrices. You will learn the general theory about matrices in linearalgebra (topics in your paper Mathematics I).First of all, we have elementary operations among 2 2 matrices: if

A =

a11 a12a21 a22

, B =

b11 b12b21 b22

then we can form a matrix A + B by simply adding their corresponding entries

A + B =

a11 + b11 a12 + b12a21 + b21 a22 + b22

.

If is a number we may form a matrix

A =

a11 a12a21 a22

.

That is A B = (aij bij) and A = (aij). The more interesting operation is the multiplicationof two matrices, defined as the following

AB =

a11 a12a21 a22

b11 b12b21 b22

=

a11b11 + a12b21 a11b12 + a12b22a21b11 + a22b21 a21b12 + a22b22

.

That is, if AB = (cij) then the entry

cij = (ai1, ai2)

b1jb2j

= dot product of (ai1, ai2) and (b1j, b2j).

Example 4.1 Let

A =

2 11 3

and B =

0 32 5

Find A + B, A B, A, AB and BA.

In general we have A + B = B + A, C(A + B) = AC + CB , (AB)C = A(BC), but themultiplication of matrices is in general not commutative.

The determinate of a 2 2 matrix A is denoted by det(A) or |A| defined by

det(A) =

a11 a12a21 a22 = a11a22 a12a21.

The mapping A |A| is not additive, but it is multiplicative i.e.

det(AB) = det(A)det(B) = det(BA).

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We will use I to denote the identity matrix

1 00 1

.

It is trivial that IA = AI for any 2 2 matrix. Clearly |I| = 1.Given a 2 2 matrix A, we say a 2 2 matrix B (if ever exists) is an inverse matrix of A if

AB = BA = I. Since det(AB) = det(A)det(B), so that a necessary condition for the existence ofan inverse matrix to A is that det(A) = 0. It turns out this condition is also sufficient.Theorem 4.2 Let A = (aij) be a 2 2 matrix. Then A has an inverse matrix if and only ifdet(A) = 0. In this case, the inverse matrix is unique and thus is denoted by A1, given by

A1 =1

det(A) a22 a12

a21 a11

.

Proof. By a direct computation we can see that A1 defined as above is an inverse matrix. IfB is an inverse of A, then

B = B(AA1) = (BA)A1 = IA1 = A1

so the inverse matrix is unique.

We observe that det(A) = 0, i.e. a11a22 = a21a12 means two row vectors

a11a21

and

a21a22

are proportional, that is, they are linearly dependent.

Let us consider R2 as the vector space of row vectors v =

v1v2

(also consider as 21 matrix).

Let A = (aij) be a 2 2 matrix. Then we associate A a linear mapping from R2

R2

denoted byA and defined by

Av =

a11 a12a21 a22

v1v2

=

a11v1 + a12v2a21v1 + a22v2

.

Proposition 4.3 Let A = (aij) be a 2 2 matrix.1) The linear system Av = 0 has no zero solutions if and only if det(A) = 0.2) The linear system Av = v has no trivial solution v = 0 if and only if is an eigenvalue

of A, that is, det(A I) = 0 (which is called the characteristic equation of A). In this case, v iscalled an eigenvector (corresponding to the eigenvalue ).

A square matrix A = (aij) is diagonal if aij = 0 for any i

= j.

Theorem 4.4 Suppose a 2 2 matrix A = (aij) has distinct real eigenvalues 1 and 2, and letvi =

v1iv2i

be corresponding vectors with eigenvalues i (i = 1, 2). Let

P =v1 v2

=

v11 v12v21 v22

.

Then

P1AP =

1 00 2

.

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Proof. First show that P is invertible, which is equivalent to that v1 and v2 are linearlyindependent. Suppose v1 + v2 = 0, so that Av1 + Av2 = 0. Hence 1v1 + 2v2 = 0. It

follows that (2 1)v

2 = 0, so that = 0 and similarly = 0. Thereforev

1 andv

2 are linearlyindependent, and P1 exists.By definition

AP = Av1 v2

=

Av1 Av2

=

1v1 2v2

= P

1 00 2

.

Since P1 exists so that

P1AP =

1 00 2

.

Example 4.5 Find all the eigenvalues and eigenvectors for the following matrices

A =

2 16 3

, B =

2 10 2

and C =

0 11 0

.

5 Systems of linear differential equations

Consider a linear differential equation of order n:

y(n) + an1y

(n1) +

+ a0y = f(t).

By introducing functions yk = y(k) where k = 0, , n 1, the previous linear equation of order

n is equivalent to the following system of linear equations of first order:

yn1 = an1yn1 a0y0 + f(t),yn2 = yn1, y0 = y1.

For example, a second order linear differential equation

d2ydt2

+ a dydt

+ by = f(t)

is equivalent to the system dxdt = ax by + f(t)dydt = x.

In terms of matrix notations, it can be written as dxdtdydt

=

a b1 0

x

y

+

f(t)

0

.

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Example 5.1 Solve the following initial value problem

dxdt

= 3x + y,dydt = 6x + 4y,

x(0) = 1;y(0) = 1.

Method 1. From the first equation, substitute y = dxdt 3x to the second equation, to obtain

d2x

dt2 3 dx

dt= 6x + 4

dx

dt 12x

so x solves the homogenous linear equation of second order

d2x

dt2 7 dx

dt+ 6x = 0

whose auxiliary equation has roots 1 and 6, so x(t) = C1et + C2e6t. Since x(0) = 1 and x(0) =y(0) + 3x(0) = 4 we have

C1 + C2 = 1, C1 + 6C2 = 4.

Method 2. By chain rule (or the invariance of first order differentials) we have

dy

dx=

dydtdxdt

=6x + 4y

3x + y

which is homogenous. By substitution u = yx , then y = xu, y = u + xu, so that

u + xdu

dx=

6 + 4u

3 + u

which is separable.We next describe another method which is contained in the following

Theorem 5.2 Consider the system of linear equations with constant coefficientsdxdtdydt

=

a11 a12a21 a22

x

y

.

Suppose A = (aij) has two distinct eigenvalues k with corresponding eigenvectors vk = v1kv2k

(k = 1, 2). Then the general solution of the system is given byx(t)y(t)

= C1e

1tv1 + C2e

2tv1

where C1 and C2 are arbitrary constants.

Proof. Let P = ( v1 v2 ). Then we have shown that

P1AP =

1 00 2

.

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Let

z(t) = z1(t)z2(t) = P

1

x(t)y(t) .

Then

d

dtz(t) = P1

ddtx(t)ddty(t)

= P1A

x(t)y(t)

= P1APz(t) =

1 00 2

z1(t)z2(t)

.

That isz1(t) = 1z1(t) and z

2(t) = 2z2(t)

so that zk(t) = Ckekt (k = 1, 2). Hence

x(t)y(t)

= P

z1(t)z2(t)

= ( v1 v2 )

z1(t)z2(t)

= C1e1t

v1 + C2e2t

v2.

Remark 5.3 If 1 = + i is complex (where = 0) andv = v1 +v2i is a (complex) eigenvector,then the general solution is given by

x(t)y(t)

= etC1 (v1 cos t v2 sin t)

+etC2 (v2 cos t + v1 sin t)

where C1, C2 are arbitrary constants.

Example 5.4 Solve the system of linear differential equationsdxdtdydt

=

0 12 3

x

y

.

Solve the characteristic equation det(A I) = 0, i.e. 2 3 + 2 = 0 to obtain eigenvalues1 = 1 and 2 = 2. For 1 = 1, solve the linear system

0 1 12 3 1

c1c2

= 0

to obtain c1 = c2, so

11

is an eigenvector with eigenvalue 1. Similarly, solve the linear system

0 2 12 3 2

c1c2

= 0

to obtain an eigenvector

12

, so the general solution

x(t)y(t)

= C1e

t

11

+ C2e

2t

12

.

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Example 5.5 Solve the system

dxdtdydt

= 2 52 4 xy .

The characteristic equation of the matrix in system 2 52 4 = 2 + 2 + 2 = 0

has a pair conjugate complex roots 1 = 1 + i and 2 = 1 i. For 1 = 1 + i the linear system2 1 5

2 4 1

c1c2

= 0

has a solution vector 53

=

53

+

01

i

thus x(t)y(t)

= etC1

53

cos t

01

sin t

+C2

01

cos t +

53

sin t

.

Example 5.6 Solve the initial value problem to the linear system dxdtdydt

=

2 14 6

x

y

,

x(0) = 1,y(0) = 1.

The characteristic equation of the matrix in the system 2 14 6 = ( 4)2 = 0

has repeated root 4, so e4t is a solution to the system. Taking into account the initial condition,we may set x(t) = (At + 1)e4t and y(t) = (Bt + 1)e4t, and feed them into the system to obtainA =

1 and B =

2. Thus the solution to the initial problem is given by

x(t) = (1 t)e4t,y(t) = (1 2t)e4t.

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Prelims Introductory Calculus 2012MT