mathematical method

8/17/2019 mathematical method

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1 Introduction

• In the sciences and engineering, mathematical models are developed to aid in the understandingof physical phenomena.

• These models often yield an equation that contains some derivatives of an unknown function.

• Such an equation is called a ’differential equation ’.

• The simplest example of model developed in calculus is the free fall of a body.

# In this case, an object is dropped from a certain height above the ground and falls under

the force of gravity.

# Newton’s second law, which states that object’s mass tines its acceleration equal the total

force acting on it, can be applied to the falling object. This leads to the equation

md2h

dt2 = −mg

# This is a differential or equation containing the second derivative of the unknown height,

h as a function of time.

# Fortunately, the above equation is easy to solve h. (Divide the equation by m)

md2h

dt2 = −mg

d2h

dt2 = −g

dh

dt = −gt + c1h(t) = −gt

2

2 + c1t + c2

• Diferential equation arise a variety of subject areas, including not only the physical sciencesbut also such diverse fields as economics, medicine, psychology and operations research. For

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Examples :

1. A classical application of differential equations is found in the study of an electric circuit.

L is the inductance

R is the resistance

E (t) is the electromotive force

q (t) is the charge

t is the time

The application of Kirchoff’s laws leads to the equation

Ld2q

dt2 + R

dq

dt +

1

cq = E (t)

2. In psycology, one model of the learning of a task involves the equationdydt

y32 (1 − y) 32 =

2 p√ n

Here the variable y represents the learner’s skill level as a function of time t. The constants

p and n depend on the individual learner and the nature of the task.

2 Terminology

• If an equation involves the derivative of ane variable with respect to another, then the formeris called a dependent variable and the latter an independent variable . Thus in the equations :

d2x

dt2 + a

dx

dt + bx = 0 (1)

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t is the independent variable

x is the dependent variable

a, b is the coefficients

∂u

∂x − ∂u

∂y = x − 2y (2)

x, y is the independent variable

u is the dependent variable

• A differential equation involving ordinary derivatives with respect to a single independentvariable is called an ordinary differential equation . (exp : eq. 1)

• A differential equation involving partial derivatives with respect to more than one independentvariable is called an partial differential equation . (exp : eq. 2)

• The order of a differential equation is the order of the highest-order derivatives present in theequation.

Examples :

1. d3u

dt3 + 7 du

dt + u = sint

2. 5 d2x

dy2 + 7

d3xdy3

5+ 2x = 0

3.

d4 pdt4

2+ d

3 pdt3

= 6t

• The degree of a differential equation is the degree of the highest-order derivatives present inthe equation.

Examples :

1.

2.

3.

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• A linear differential equation is one in which the occurrence of the dependent variable y hasthe format

an(x)dny

dxn + an−1(x)

dn−1ydxn−1

+ · · · + a1 dydx

+ a0(x)y = F (x)

where

1. an(x), an−1(x), . . . , a0(x) and F (x) depend only on the independent variable x.

2. the highest degree of the dependent variable y and its derivative is up to 1.

3. there’s no multiplication between y and any of its derivatives.

4. there’s no transcendental function of y or its derivatives.

Examples :

1. d2y

dx2 + y3 = 0

2. d2y

dx2 − y3 = x3

3. d2ydx2

+ y dydx

= cosx

4. y′

+ 1x

y = x3 − 3

5. x′

+ 3y

x = 2y

6. dydx

= a(x)y + b(x)

7. (1 + y2

)d2y

dt2 + tdy

dt + y = ey

3 Solution and Initial Value Problems

• An nth order ordinary differential equation is an equality involving the independent variablex, the dependent variable y, and the first n derivatives of y.

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Examples :

x2d2y

dx2x

dy

dx + y

= x3

(2

nd

order)

√ 1 −

d3y

dx3

2− y = 0 (3rdorder)

d4y

dx4 = xy (4thorder)

• Thus a general form for an nth order equation would be

F

x,y,

dy

dx, . . . ,

dny

dxn

= 0 (3)

where,

# F is a function of the independent variable x, the dependent variable y, and the derivatives

of y up to order n; that is, x, y , . . . , dny

dxn.

# We assume that the equation holds for all x in an interval I ; [a, b] or (a, b] or [a, b) or

(a, b).

• In many cases, we can isolate the highest-order term dnydxn

, and write as

dny

dxn = f

x,y,

dy

dx, · · · , d

n−1ydxn−1

(4)

4 Explicit Solution

A function φ(x) that when subtituted for y in equation (3) and (4) satisfies the equation for all x

in the interval I is called an explicit solution to the equation on I .

Example 1:

Show that φ(x) = x2 − x−1 is an explicit solution to

d2y

dx2 − 2

x2y = 0 (5)

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Solution :

φ(x) = x2

− x−1

φ′

(x) = 2x + x−2

φ′′

(x) = 2 − 2x−3

Substitute φ′

(x) and φ′′

(x) into (5),

(2 − 2x−3) − 2x2

(x2 − x−1)

= 2 − 2x−3) − 2x2

(2 − 2x−3)= 0

So, φ(x) = x2 − x−1 is an explicit solution to (5) on (−∞, 0) ∪ (0, ∞).

Example 2 :

Show that for any choice of the constants c1 and c2, the function φ(x) = c1e−x + c2e2x is an

explicit solution to

y′′ − y′ − 2y = 0 (6)

Solution :

φ(x) = c1e−x + c2e2

x

φ′

(x) = −c1e−x + 2c2e2xφ′′

(x) = c1e−x + 4c2e2

x

Substitute φ(x), φ′

(x) and φ′′

(x) into (6),

(c1e−x + 4c2e2

x) − (−c1e−x + 2c2e2x) − 2(c1e−x + c2e2x)= (c1 + c1 − 2c1)e−x + (4c2 − 2c2 − 2c2)e2x= 0

So, φ(x) = c1

e−x + c2

e2x is an explicit solution to (6) on the interval (−∞

,∞

) for any choice

of the constants c1 and c2.

5 Implicit Solution

A relation G(x, y) = 0 is said to be an implicit solution to equation (3) on the interval I if it

defines one or more explicit solutions on I .

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Example 1:

Show that y2

− x3

+ 8 = 0 is an implicit solution tody

dx =

3x2

2y (7)

on the interval (2, ∞).Solution :

y2 − x3 + 8 = 0y = ±√ x3 + 8

Both φ(x) and φ′

are defined on (2, ∞). Substitute them into (7),

3x2

2√

x3 + 8=

3x2

2(√

x3 + 8)

Then, y2 − x3 + 8 = 0 is an implicit solution to (7) on the interval (2, ∞).(You can check that φ(x) = −√ x3 + 8 is also an implicit solution to (7))-Note : We hereafter use the term ’solution ’ to mean an explicit or implicit solution.

5.1 Initial Value Problem

• By an ’initial value problem ’ for an nth order differential equation

F

x,y,

dy

dx, . . . ,

dny

dxn

= 0

We mean : Find a solution to the differential equation on an interval I that satisfies at x0 the

n initial conditions.

y(x0) = y0dy

dx(x0) = y1

...dn−1ydxn−1

(x0) = yn−1

where x0 ∈ I and y0, y1, . . . , yn−1 are given constant.

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• In the case of a first order equation, the initial conditions reduce to the single requirementy(x0) = y0

and in the case of a second order equation , the initial conditions have the form

y(x0) = y0, dy

dx(x0) = y1

Example 1:

Show that φ(x) = sinx − cosx is a solution to the initial value problem

d2y

dx2 + y = 0 (8)

where y(0) = −1, dydx

(0) = 1

Solution :

φ(x) = sinx − cosxφ′

(x) = cosx + sinx

φ′′

(x) = −sinx + cosx

Substitute φ(x) and φ′′

(x) into (8),d2y

dx2 + y

= (−sinx + cosx) + (sinx − cosx)= 0

So, φ(x) is a solution to (8) on the interval (−∞, ∞) .

When we check the initail conditions, we find

φ(0) = y(0) = sin0 − cos0 = −1φ′

(0) = dy

dx(0) = −sin0 + cos0 = 1

which meets the requirements of (8). Therefore φ(x) is a solution to the given initial value

problem.

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Example 2 :

Given the general solution for the initial value problem :y′′ − y′ − 2y = 0, y(0) = 2, y′(0) = −3

is y(x) = c1e−x + c2e2x. Find the value of c1 and c2.

Solution :

y(x) = c1e−x + c2e2

x

y′

(x) = −c1e−x + 2c2e2x

y(0) = 2 : c1e−(0) + c2e2(0) = 2

c1 + c2 = 2 (9)

y′

(0) = −3 : −c1e(0) + 2c2e2(0) = −3−c1 + 2c2 = −3 (10)

(9) + (10) : c2 = 13

, c1 = 73

So, y(x) = 73

e−x − 13

e2x ⇒ particular solution

6 Separable Equation

dy

dx = f (x, y)

dy

dx =

depends only onx g(x) · h(y)

depends only ony∫ 1

h(y)dy =

∫ g(x)dx + c

Example 1:

dx

dt = 3xt2∫

1

3xdx =

∫ t2dt

1

3ln|x| = t

3

3 + c

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Example 2 :

dy

dx =

yex+y

x2 + 2dy

dx =

yexey

x2 + 2∫ 1

yeydx =

∫ ex

x2 + 2dt

Example 3 :

Solve the initial value problem :

dy

dx

= y − 1x + 3

, y(

−1) = 0

Solution :

This is the separable equation, so∫ 1

y − 1dy =∫

1

x + 3dx

ln|y − 1| = ln|x + 3| + c (11)Then use the initial value, y(−1) = 0 . Substitute x = −1, y = 0 into (11) we get,

ln|0 − 1| = ln| − 1 + 3| + cln1 = ln2 + c

0 = ln2 + c

c = −ln2So, (11) becomes,

ln|y − 1| = ln|x + 3| − ln2ln|y − 1| = ln |x + 3|

2

|y − 1| = |x + 3|2

y − 1 = (x + 3)2

or y − 1 = −(x + 3)2

y = −(x + 5)2

it’s not satisfied the initial value problem, becausey(−1)= (−1+5)2 =2̸=0

y = −(x + 1)2

satisfied the initial value problem, becausey(−1)=− (−1+1)2 =0

So, the solution (explicit solution) for the initial value problem is y = −12

(x + 1).

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7 Linear First Order Equation

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Example 1:

Find the general solution to dydx

+ y = 2 + 2x.

Solution :

We can clearly see that P (x) = 1 and Q(x) = 2 + 2x.

Find the integrating factor : µ(x) = e∫ 1dx = ex

ex

dy

dx

+ (−exy) = e

x(2 + 2x)

d

dx[ex · y] = ex(2 + 2x)

ex · y = ∫ ex(2 + 2x)dxy =

?

ex +

c

ex

y = ?

ex + A, A =

c

ex.

Example 2 :

Find the general solution to 1x

dydx

− 2yx2

= x cos x, x > 0.

Solution : dy

dx − 2y

x = x2cos x ⇒ P (x) = −2

x and Q(x) = x2cos x

µ(x) = e∫ − 2

xdx = e−2ln|x|

= eln|x|−2

= |x|−2= x−2

x−2dy

dx −

2x−3y = cos xd

dx[x−2 · y] = cos x

x−2 · y =∫

cos x dx

= sin x + c

y = x2sin x + A, A = cx2.

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Example 3 :

Solve dydx

+ xy = 4x.

Solution :

Integrating factor : µ(x) = e∫

xdx = ex2

2

ex2

2dy

dx + e

x2

2 y = ex2

2 (4x)

d

dx[e

x2

2 · y] = 4ex2

2 x

ex2

2 · y =∫

4xex2

2 dx

ex22 · y = 4ex22 + cy = 4 + ce

x2

2

y = 4 + A, A = cex2

2 .

Example 4:

Solve dudv

+ 1v

u = v3, v > 0.

Solution :

Integrating factor :µ(x) = e

∫ 1v

dx

= eln

|v

| = eln v

= v, if v > 0

vdu

dv + v

1

vu

= v(v3)

vdu

dv + u = v4

d

dv[u · v] = v4

u · v =∫

v4 dv

u = v4

5 + c

v.

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8 Bernoulli Equation

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Example 1:

Solve

dy

dx − 1

2xy = 5x2y5 (12)

Solution :

(1) Divide (12) by y5.

1

y5dy

dx − 1

y51

2xy =

1

y55x2y5

y−5dy

dx − y

−4

2x = 5x2 (13)

(2) Substitute → 1. v = y1−n→ 2. ( 1

1−n)

dvdx

= y−n dydx

into (13)

n = 5 → v = y−4dv

dx = −4y−5 dy

dx1

4

dv

dx =

dy

dx

14

dvdx

− v2x

= 5x2 ⇒ Linear Differential Equation↓

change to standart form :

dv

dx +

2

xv = −20x2

(3) Use the method for solving linear equation.

Integrating factor :µ(x) = e∫

1x

dx

= e2ln|x|

= eln|x|2

= x2

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x2dvdx + x22vx = x2(−20x2)

x2dv

dx + 2vx = −20x4

d

dx[x2 · v] = −20x4

x2 · v = −∫

20x4dx

x2 · v = −20x5

5 + c

v = −4x3 + cx2

y−4 = −4x3 + A, A = cx2

Example 2 :

Solvedy

dx = xy2 + 2xy (14)

Solution :

dy

dx − 2xy = xy2 (15)

(1) Divide (14) by y2.

y−2dy

dx − 2xy−1 = x

(2) Substitute → 1. v = y1−n

→ 2. ( 11−n)

dvdx

= y−n dydx

into (15)

n = 2 → v = y−1dv

dx = (−1)y−2 dy

dx

−dvdx

= y−2dy

dx

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− dvdx

− 2xv = x ⇒ Linear Differential Equation

↓change to standart form :

dv

dx + 2xv = −x

(3) Use the method of solving linear equation.

Integrating factor :µ(x) = e∫ 2xdx

= ex2

ex2 dv

dx + 2xvex

2

= −xex2

d

dx[ex

2 · v] = −xex2

ex2 · v =

∫ −xex2 dx

ex2 · v = −1

2ex

2

+ c

v = −

1

2 +

c

ex2

y−1 = −12

+ A, A = c

ex2

9 Exact Equation

dy

dx = f (x, y)

⇓can be rewritten in the differential form :

M (x, y)dx + N (x, y)dy = 0 (16)

⇓If the left hand side of equation (16) can be identified as a total differential :

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

∂F

∂x dx +

∂F

∂y dy

= dF (x, y)

⇓then its solutions are given (implicitly) by

F (x, y) = C

for an arbitrary constant C. So, the differential form,

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

is said to be e xact in a rectangle R if there is a function F (x, y) such that

∂F (x, y)

∂x = M (x, y) and

∂F (x, y)

∂y = N (x, y)

for all (x, y) in R. That is the total differential of F (x, y) satisfies

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

⇓If M (x, y)dx + N (x, y) is an exact differential form, then the equation

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

is called an e xact equation

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9.1 Theorem 1 : Test For Exactness

Suppose the first partial derivatives of M

(x, y

) and N

(x, y

) are continuous in a rectangle R

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

is an exact equation in R if and only if the compatibility condition

∂M

∂y (x, y) =

∂N

∂x (x, y)

holds for all (x, y) in R.

Example 1:

Determine the exactness of the differential equation :dy

dx =

3x2 − yx − 1

Solution :

(x − 1)dy = (3x2 − y)dx−(3x2 − y)dx + (x − 1)dy = 0

(y − 3x2)

M (x,y)

dx + (x − 1)

N (x,y)

dy = 0 (17)

Exact or not exact ....?

Method-1:

1. Rearrange equation (17) and find F (x, y)

ydx − 3x2dx + xdy − dy = 0(ydx + xdy) − 3x2dx − dy = 0

d(xy) − d(x3) − d(y) = 0d(xy − x3 − y ) = 0

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2. Determine whether ∂F (x,y)∂x

= M (x, y) and ∂F (x,y)∂y

= N (x, y)

∂F (x, y)

∂x = y − 3x2 = M (x, y)

and∂F (x, y)

∂y = x − 1 = N (x, y)

So, (17) is an exact equation.

Method-2:

Use Theorem 1 (test for exacness), determine whether ∂M

∂y (x, y) = ∂N

∂x (x, y)

∂M

∂y (x, y) = 1 =

∂N

∂x (x, y)

So, (17) is an exact equation.

9.2 How to Solve an Exact Equation

Determine whether it is an exact equation or not, if YES

⇓Then

∃ ∂F (x, y)∂x

= M (x, y) (18)

and ∂F (x, y)

∂y = N (x, y) (19)

⇓Integrate (18) with respect to x :

F (x, y) =

∫ M (x, y) ∂x + g(y) (20)

⇓Take the partial derivative with respect to y of both sides of equation (20).

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∂F (x, y)

∂y =

∂

∂y ∫ M (x, y) ∂x + ∂ ∂y g(y g′(y)

) (21)

⇓Compare (21) with (19) and substitute N for ∂F (x,y)

∂y into (21).

⇓You’ll get g

′

(y) =?

Then it follows by g(y) =?

⇓Substitute g(y) into (20) gives F (x, y)

⇓The solution : F (x, y) = C .

Example 1:

Solve the differential equation

(y − 3x2)dx + (x − 1)dy = 0Solution :

(y − 3x2) M (x,y)

dx + (x − 1) N (x,y)

dy = 0

∂M

∂y (x, y) = 1,

∂N

∂x (x, y) = 1

⇒ So, Exact.

⇓Then

∃ ∂F (x, y)∂x

= M (x, y) = y − 3x2 (22)

and ∂F (x, y)

∂y = N (x, y) = x − 1 (23)

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Choose (22) and integrate it with respect to x :

F (x, y) = ∫ y − 3x2 ∂x + g(y)= yx − x3 + g(y) (24)

Take the partial derivative with respect to y of both sides of equation (24).

∂F (x, y)

∂y = x + g

′

(y) (25)

⇓Compare (25) with (23) and substitute N for ∂F (x,y)

∂y into (25).

⇓You’ll get g

′

(y) = −1Then it follows by g(y) = −y

⇓Substitute g(y) = −y into (24) gives F (x, y) = yx − x3 − y

⇓The solution : yx − x3 − y = C .

Example 2 :

Solve the differential equation

(2xy − sek2x)dx + (x2 + 2y)dy = 0 (26)

Solution : ∂M

∂y = 2x,

∂M

∂y = 2x ⇒ So,(26) is an exact equation

∃ ∂F (x, y)

∂x = M (x, y) = 2xy

−sek2 x (27)

and ∂F (x, y)

∂y = N (x, y) = x2 + 2y (28)

Choose (27),

F (x, y) =

∫ 2xy − sek2 x ∂x + g(y)

= x2y − tan x + g(y) (29)

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∂F (x, y)

∂y = x2

+ g′

(y) (30)

Compare (30) with (28)g′

(y) = 2y

g(y) = y2

Subsitute g(y) = y2 into (29),

So,F (x, y) = x2y − tan x + y2

x2y

−tan x + y2 = C.

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9.3 How to Determine whether it is an exact eq or not?

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Example 1:

Show that µ(x, y) = xy2 is an integrating factor for

(2y − 6x)dx + (3x − 4x2y−1)dy = 0 (31)

Use this integrating factor to solve the equation.

Solution :

(2y − 6x) M (x,y)

dx + (3x − 4x2y−1) N (x,y)

dy = 0

∂M

∂y = 2,

∂M

∂y = 3 − 8xy−1 ⇒ So,(31) is not an exact equation

Multiply (31) by the given integrating factor, µ(x, y) = xy2

(xy2)(2y − 6x)dx + (xy2)(3x − 4x2y−1)dy = 0(2xy3 − 6x2y2)dx + (3x2y2 − 4x3y)dy = 0 (32)

∂ (µM )

∂y = 6xy2 − 12x2y = ∂ (µN )

∂x ⇒ Exact

So, µ(x, y) = xy2 is an integrating factor for (31).

Now, solve (32) using exact method.

∃ ∂F (x, y)∂x

= µM (x, y) = 2xy3 − 6x2y2 (33)

and ∂F (x, y)

∂y = µN (x, y) = 3x2y2 − 4x3y (34)

F (x, y) =

∫ 2xy3 − 6x2y2 ∂x + g(y)

= x2y3 − 2x3y2 + g(y).

∂F (x, y)

∂y

= 3x2y2

−4x3y + g

′

(y)

So, g′

(y) = 0 ⇒ g(y) = 0F (x, y) = x2y3 − 2x3y2

x2y3 − 2x3y2 = C.

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Example 2 :

Show that µ(x, y) = 11+x2y2

is an integrating factor for

(1 + x2y2 + y)dx + xdy = 0 (35)

Use this integrating factor to solve the equation.

Solution :

(1 + x2y2 + y) M (x,y)

dx + x N (x,y)

dy = 0

∂M

∂y

= 2x2y + 1, ∂M

∂y

= 1

⇒ So,(35) is not an exact equation

( 1

1 + x2y2)(1 + x2y2 + y)dx + (

1

1 + x2y2)xdy = 0

(1 + y

1 + x2y2)dx + (

x

1 + x2y2)dy = 0 (36)

∂ (µM )

∂y =

1 − x2y2(1 + x2y2)2

= ∂ (µN )

∂x ⇒ Exact

So, µ(x, y) = 11+x2y2

is an integrating factor for (35).

Now, solve (36) using exact method.

∃ ∂F (x, y)∂x

= µM (x, y) = 1 + y1 + x2y2

(37)

and ∂F (x, y)

∂y = µN (x, y) =

x

1 + x2y2 (38)

F (x, y) =

∫ 1 +

y

1 + x2y2 ∂x + g(y)

= x +

∫ 1

1 + (xy)2 ∂ (xy) + g(y)

= x + tan−1(xy) + g(y).

∂F (x, y)

∂y = 0 +

x

1 + (xy)2 + g

′

(y)

So, g′

(y) = 0 ⇒ g(y) = 0F (x, y) = x + tan−1(xy)

x + tan−1(xy) = C.

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9.4 Theorem 2 : Special Integrating Factor

• If 1

N ∂M ∂y − ∂N ∂x is continuous and depends only on x, thenµ(x) = exp

∫ { ∂M ∂y

− ∂N ∂x

N

}dx

is an integrating factor for M (x, y)dx + N (x, y)dy = 0

• If 1M

∂N ∂x

− ∂M ∂y

is continous and depends only on y, then

µ(y) = exp

∫ { ∂N ∂x

− ∂M ∂y

M

}dy

is an integrating factor for M (x, y)dx + N (x, y)dy = 0

Example 1:

Solve

(2x2 + y)dx + (x2y − x)dy = 0. (39)

(1) Test the exactness

∂M ∂y

= 1, ∂N ∂x

= 2xy

−1

⇒ not exact

(2) Compute 1N

∂M ∂y

− ∂N ∂x

1

x2y − x

[1 − (2xy − 1)]

= 2 − 2xy

x2y − x=

2(1 − xy)

−x(1

−xy)

= −2x

= f (x)

µ(x) = e∫ − 2

xdx = e−2ln |x| = |x|−2 = x−2

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(3) Multiply (39) by 4µ(x) = x−2

x−2(2x2 + y)dx + x−2(x2y

−x)dy = 0

(2 + yx−2) µM

dx + (y − x−1) µN

dy = 0

(4) Exact / not?∂ (µM )

∂y = x−2 =

∂ (µN )

∂x ⇒ Exact

...

F (x, y) = 2x − yx−1 + y2

2 = C.

Example 2 :

Solve(2xy)dx − (x2 + y2 − 1)dy = 0. (40)

∂M ∂y

= 2x, ∂N ∂x

= −2x ⇒ not exact

1

N

∂M

∂y − ∂N

∂x

= 1x2 + y2 − 1 [2x − (−2x)]=

4x

x2 + y2 − 1= f (x)

1

M

∂N

∂x − ∂M

∂y

= 1

2xy [−

2x

− −2x]

= 4x

x2 + y2 − 1= −−4x

2xy

= −2y

= g(y) ⇒ easier

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µ(x) = e∫ − 2

ydy = e−2ln |y| = |y|−2 = y−2

(4) Exact / not? ∂ (µM )

∂y = −2xy−2 = ∂ (µN )

∂x ⇒ Exact

...

F (x, y) = x2y−1 − y − 1y

= C.

10 Homogeneous Equations

10.1 Homogeneous

A function f (x, y) is called a homogeneous with degree of n if

f (λx,λy) = λnf (x, y)for all λ.

Example 1:

f (x, y) = x4 − x3y

f (λx, λy) = (λx)4 − (λx)3(λy)= λ4x4 − λ4x3y= λ4[x4 − x3y]= λ4f (x, y)

So, f (x, y) is a homogeneous with degree of 4.

Example 2 :

(y−xexy )

x

f (λx, λy) = (λy) − (λx)e

(λx)(λy)

(λx)

= λ(y − xeλ(x)λ(y) )

λ(x)

= λ0f (x, y)

So, f (x, y) is a homogeneous with degree of 0.

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10.2 Theorem 3

If M

(x, y

) and N

(x, y

) homogenous with the same degree,a function f (x, y) = M (x,y)

N (x,y)

is homogeneous with degree of zero.

Proof

If M (x, y) and N (x, y) homogeneous with degree n,

so, M (λx, λy) = λnM (x, y)

and, N (λx, λy) = λnN (x, y)

f (x, y) = M (λx, λy)

N (λx, λy)

= λnM (x, y)

λn

N (x, y)

= λ0f (x, y)

10.3 Theorem 4

If a function f (x, y) is hommogeneous with degree of zero, so f (x, y) can be expressed as a function

of the ratio yx

alone.

Proof

Let v = yx

⇒ vx = y

f (x, y) = f (x,vx) y = vx

= x0f (1, v) f (x, y)homogeneous with degree of zero

= f (1, v)

= a function that depends only on v.

Example 1:M (x, y)dx + N (x, y)dy = 0

N (x, y)dy = −M (x, y)dxdy

dx = −M (x, y)dx

N (x, y)dy M and N homogeneous with the same degree

= f (x, y) from Theorem 3

= F (y

x) from Theorem 4 (where v =

y

x)

v = y

x ⇒ y = vx

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dy

dx v +

dv

dx x = F (v)x

dx =

F (v) − vdv∫

1

F (v) − v dv =∫

1

x

10.4 How to solve a homogeneous equation?

M (x, y)dx + N (x, y)dy = 0 (41)

⇓is called a homogeneous equation if both M (x, y) and N (x, y) are homogeneous with the same

degree.

⇓Solution : Substitution of y = vx

⇒ dy = xdv + vdx

⇓becomes : separable equation

⇓integrate

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Example 1:

Solve

(x − y)dx + xdy = 0 (42)

M (λx, λy) = (λx) − (λy)= λ1(x − y)

N (λx, λy) = (λx)

= λ1(x)

From the definition, if M and N homogeneous with the same degree, so (42) is a homogeneous

equation.

v = y

x ⇒ y = vx

(i)

⇒ dydx

= v + xdv

dx

dy = vdx + xdv (ii)

Substitute (i) and (ii) into (42) :

(x − (vx))dx + x(vdx + xdv) = 0(x

−vx + vx)dx + x2dv = 0

xdx + x2dv = 0

x2dv = −xdx∫ dv =

∫ −1

x

v = −ln|x| + c.

Evercise : Show that (42) also can be solve by using Exact Method.

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Example 2 :

Solve

(xy − y2 + x2)dx − x2dy = 0 (43)

M (λx, λy) = (λx)(λy) + (λy)2 + (λx)2

= λ2(xy) + λ2y2 + λ2x2

= λ2(xy + y2 + x2)

= λ2M (x, y) ⇒ Hom. with degree 2

N (λx, λy) = −(λx)2

= −λ2x2= λ2(−x2)= λ2N (x, y) ⇒ Hom. with degree 2

From the definition, if M and N homogeneous with the same degree, so (43) is a homogeneous

equation.

v = y

x ⇒ y = vx

(i)

⇒ dydx

= v + xdv

dx

dy = vdx + xdv (ii)

Substitute (i) and (ii) into (43) :

(x(vx) + (vx)2 + x2)dx − x2(vdx + xdv) = 0(vx2 + v2x2 + x2 − vx2)dx − x3dv = 0

(v2x2 + x2)dx − x3dv = 0x2(v2 + 1)dx − x3dv = 0

x2(v2 + 1)dx = x3dv∫ 1x

dx =∫ 1

v2 + 1dv

c + ln |x| = tan−1(v)ln |x| + c = tan−1( y

x)

y

x = tan (ln |x| + c)

y = x tan (ln |x| + c).

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Example 1:

Solve

(x − 2y + 1)dx + (4x − 3y − 6)dy = 0 (44)

a2

a1=

4

1 = 4,

b2

b1=

−3−2 =

3

2 ⇒ Case I

(h, k) satisfies :

h − 2k + 1 = 04h − 3k − 6 = 0

⇒ h = 3, k = 2

Substitute : x = X + 3 and y = Y + 2 into (44)

[(X + 3) − (Y + 2) + 1]dX + [4(X + 3) − 3(Y + 2) − 6]dY = 0(X − 2Y )dX + (4X − 3Y )dY = 0 (45)

Substitute Y = vX and dY = vdX + Xdv into (45)

(X − 2vX )dX + (4X − 3vX )(vdX + Xdv) = 0(X − 2vX + 4vX − 3v2X )dX + (4X 2 − 3vX 2)dv = 0

(X + 2vX − 3v2X )dX + (4 − 3v)X 2dv = 0(1 + 2v − 3v2)dX + (4 − 3v)Xdv = 0

(4 − 3v)Xdv = −(1 + 2v − 3v2)dX ∫ 4 − 3v1 + 2v − 3v2dv = −

∫ 1

X dX

...

X 4|3v + 1|5 = |v − 1|A, A = e4cX 4|3 Y

X + 1|5 = | Y

X − 1|A, ⇒ v = Y

X

|3Y + X |5 = |Y − X |A|3(y − 2) + (x − 3)|5 = |(y − 2) − (x − 3)|A, ⇒ X = x − 3, Y = y − 2

|3y + x − 9|5 = A|y − x + 1|.

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Example 2 :

Solve

(x + 2y + 3)dx + (2x − 3y − 6)dy = 0 (46)

a2

a1=

2

1 = 2,

b2

b1=

4

2 = 2 ⇒ Case II

Let

z = a1x + b1y ⇒ z = x + 2y

(i)⇒ dz

dx = 1 + 2

dy

dx

dz = dx + 2dy

dy = dz − dx

2 (ii)

(46) becomes :

(z + 3)dx + (2z − 1)( dz − dx2

) = 0

(2z + 6)dx + (2z − 1)(dz − dx) = 0(2z + 6

−2z + 1)dx + (2z

−1)dz = 0

7dx + (2z − 1)dz = 0∫ (2z − 1)dz =

∫ −7dx

z 2 − z = −7x + c(x + 2y)2 − (x + 2y) + 7x = c, ⇒ z = x + 2yx2 + 4xy + 4y2 + 6x − 2y = c.

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12 Mixing Problems

• A problem for which the one-compartment system provides a useful representation is themixing of fluids in a tank.

• Let x(t) represent the amount of a substance in a tank (compartment) at time, t.• To use the compartmental analysis model, we must be able to determine the rates at which

this substance enters and leaves the tank.

• In mixing problems one is often given the rate at which a fluid containing the substance flowsinto the tank, along with the concentration of the substance in that fluid.

• Hence, multiplying the flow rate (volume/time) by the concentration (amount/volume) yieldsthe input rate (amount/time).

• The output rate of the substance is usually more difficult to determine.• If we are given the exit rate of the mixture of fluids in the tank, then how do we determine

the concentration of the substance in this mixture?

• One simplifying assumption that we might make is that the concentration is kept uniform inthe mixture. Then we can compute the concentration of the substance in the mixture by

dividing the amount x(t) by the volume of the mixture in the tank at time, t.

• Multiplying this concentration by the exit rate of the mixture then gives the desired outputrate of the substance.

• The problem : What is the amount of a substance in a tank at time t (x(t) =?).

• dxdt

= changing rate

= input rate ( amounttime )kgmin−1

- output rate ( amounttime )kgmin−1

= (con. of a substance) ( amountvolume )kgL−1

· (enter rate) ( volumetime )Lmin−1

− ( con. of the subs. in the mix.) ( amount sub. in the tank(mix),x(t)volume of the mix. in the tank )kgL−1

· (exit rate) ( volumetime )Lmin−1

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Example 1:

Consider a large tank holding 1000L of water into which abrine solution of salt begins to flowat a constant rate of 6Lmin−1. The solution inside the tank is kept well stirred and is flowing

out of the tank at a rate of 6Lmin−1. If the concentration of salt in the brine entering the

tank is 1kgL−1, determine when the concentration of salt in the tank will reach concentration12

kgL−1.

dxdt

= input rate - output rate

= (con. of salt) · (enter rate) −

amount salt in the tankvolume of fluid in the tank

· (exit rate)

= (1kgL−1) · (6Lmin−1) − x(t)kg1000L · (6Lmin−1)= 6kgmin−1 − 6x

1000kgmin−1

So, dx

dt = 6 − 6x

1000, x(0) = 0

= 3000 − 3x

500

= 3(1000 − x)

500∫ 11000 − x dx =

∫ 3

500dt

−ln|1000 − x| = 3500

t + c

ln|1000 − x| = − 3t500

− c1000 − x = e− 3t500 − c1000 − x = e− 3t500 e−c

x = 1000

−Ae−

3t500 , A = e−c

x(0) = 0 : 0 = 1000 − Ae0A = 1000

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So,x(t) = 1000(1 − e− 3t

500 )= amount salt in the tank at time t

x(t)

1000 = the con. of salt in the tank at time t.

To determine when the con. of salt is 12

kgL−1, we set the right-hand side equal to 12

and solve for

t.

1 − e− 3t500 = 121

2 = e− 3t500

ln|12| = − 3t

500

ln1 − ln2 = − 3t500

t = 500 ln2

3 ≈ 115.5min.

If enter rate

(exp:6Lmin−1)̸= exit rate

(exp:6Lmin−1)dx

dt = (1kgL−1) · (6Lmin−1) −

x(t)kg

(1000 + t)L

· (5Lmin−1)

The difference between the rate of flow into the tank and the rate of flow out : 6 - 5 = 1 Lmin−1

So, the volume of fluid in the tank after t minutes is (1000 + t)L.

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13 Kinematics

• Kinematics is the science of decribing the motion of object without regard to the causes of themotion.

• In this section, we shall consider motion along a straight line, that is one-dimensional motion.

13.1 Terminology

• Scalar - are quantities which are fully described by magnitude alone.• Vector - are quantities decribed by magnitude and direction.• Displacement, X - is a vector quantity which refers to ”how far out of place an object is”; it is

the object’s change in position.

• Speed - is a scalar quantity which refers to ”how fast an object is moving.- A fast-moving object has a high speed while a slow-moving object has a low speed.

- An object with no movement at all has a zero speed.

- Average speed =

distance traveled,x

time of travel,t- Instantaneous speed = dx

dt

• Velocity - is a vector quantity which refers to ”the rate at which an object changes its position.- When evaluating the velocity of an object, one must keep track of direction.

- Average velocity = Displacement,X

time,t- Instantaneous velocity, dX

dt - the velocity of a particle at any instant of time.

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(c) Acceleration of the object within the above time

Acceleration = dv

dt =

d2x

dt2 = a

a = d2x

dt2 = 6t − 12

t = 5 ⇒ d2x

dt2 = a = 6(5) − 12 = 18ms−2

(d) Distance during time 4 ≤ t ≤ 6s

v = dxdt

= 3t2 − 12t − 15= 3(t + 1)(t − 5)

Object position

t = 4, x(4) = −52mt = 5, x(5) = −60mt = 6, x(6) = −50m

So, total distance = 8 + (8 + 2) = 18m.

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14 Vertical Motion Under Gravity

Example 1:

An object is thrown straight upwards with an initial velocity 15ms−2. Acceleration under

gravity is 9.8ms−2. Calculate :

(a) time taken to reach maximum height.

Acceleration = dv

dt = −g (Separable equation)

⇒ v = −gt + c

Initial velocity, v0 = 15ms−1

v(0) = 15 (t = 0, v = 15)

⇒ c = 15

So, v = −gt + 15 (47)

Maximum height ⇒ v = 00 = −9.8t + 15t =

15

9.8 = 1.53s.

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(b) the maximum height from the original position.

(47) can be rewritten as dxdt

= −gt + 15∫ dx =

∫ −gt + 15dt

x = −gt2

2 + 15t + c

t = 0, x = 0 ⇒ c = 0x =

−

gt2

2

+ 15t (48)

Time taken to reach maximum height, t = 1.53

So, maximum height, x = −9.8(1.53)22

+ 15(1.53) = 11.48m.

Alternatively

dv

dt =

dv

dx · dx

dt = v

dv

dx

Since dv

dt

=

−g,

So, vdv

dx = −g ⇒ Separable∫

vdv =

∫ −gdx

v2

2 = −gx + c

x = 0, v = 15 ⇒ 152

= −9.8(0) + cc = 112.5m

v2

2 = −gx + 112.5 (49)

At the maximum height,v = 0

0 = −9.8x + 112.5x = 11.48m

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(c) time taken to return to its original position.

From (48); x = gt2

2 + 15t

x = t

−gt

2 + 15

x = 0 ⇒ t = 0s and t = 3.06s.

So, time taken to return to its original position, t = 3.06s.

(d) velocity when it return to its original position.

From (47); v = −gt + 15t = 3.06 ⇒ v = −9.8(3.06) + 15

v = −14.99 ≈ −15ms−1

Obbject moves with velocity 15ms−1 downwards.

15 Newtonian Mechanics

• Mechanics is the study of the motion of objects and the effect of forces acting on those objects.

• Deals with the motion of ordinari objects, that is, object that are large compared to an atomand slow moving compared with the speed of light.

• A model for Newtonian mechanics can be based on Newton’s laws of motion :1. Let F 1, F 2 are external force.

If F 1 + F 2 = 0, then it moves with a constant velocity.

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2. When a body is subject to one or more external forces, let F i, i = 1, 2, 3

then, dpdt

= F 1 + F 2 + F 3

3. When one body interacts with a second body,

15.1 Procedure for Newtonian Models

(a) Determine all relevant forces actin on the object being studied.

(It is helpful to draw a simple diagram of the object that depects theses forces)

(b) Choose an appropriate axis or coordinate system in which to represent the motion of the

object and the forces acting on it. Keep in mind that this coordinate system must be an

inertial reference frame.

(c) Apply Newton’s second lawdp

dt = F (t,x,v), p(t) = mv(t) (50)

or dp

dt =

d(mv)

dt = m

dv

dt = ma = F (t,x,v) (51)

whichever is appropriate, to determine the equations of motion for the object.

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Example 1:

An object of mass 1 kg is released and falls under the influence of gravity. The force due to air

resistance is 7(x+1) with x being its distance from the origin. Find the velocity of the object

when it falls to x = 6.4m. [Use g = 10ms−2]

Solution :

mdv

dt = mg − 7

x + 1dv

dt = g − 7

x + 1 (if m = 1)

dv

dx · dx

dt = g − 7

x + 1dv

dx · v = g − 7

x + 1∫ vdv =

∫ g − 7

x + 1

dx

v2

2 = gx − 7ln|x + 1| + c

x = 0, v = 0

0 = 0 − 7ln1 + cc = 0

⇒ v2

2 = gx − 7ln|x + 1|

if x = 6.4m,

so, v2

2 = 10(6.4) − 7ln|7.4|

v = 9.99 ≈ 10ms−1

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Example 2 :

An object of mass mkg is given a downward velocity of v0 and is then allowed to fall under theinfluence of gravity. Assume that the influence of gravity is constant and the force due to air

resistance is proportional to the velocity of the object. Determine the equation of the motion

of the object.

Solution :

mdv

dt = mg − kv

∫ mmg − kv dv = ∫ dt

−mk

ln|mg − kv| = t + c

ln|mg − kv| = −ktm

+ ck

m

|mg − kv| = e−ktm+ ckm|mg − kv| = e−ktm A, A = ck

m

t = 0, v = v0

So, mg − kv0 = A⇒ mg − kv = e−ktm (mg − kv0)

−kv = e−ktm (mg − kv0) − mg

v = mg

k − e

−ktm

k (mg − kv0)

v(t) = mg + e−ktm

v0 − mg

k

(52)

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v(t) =

dx

dt = mg + e−kt

m v0 − mgk ∫ dx =

∫ mgk

t + e−ktm

v0 − mg

k

dt

x =mg

k

t +

e−ktm

(− km

)

v0 − mg

k

+ c

x =mg

k

t − me

−ktm

k

v0 − mg

k

+ c

t = 0, x = 0

So, 0 = 0 − me0

k

v0 − mg

k

+ c

c = m

k

v0 − mg

k

Then, x =mg

k

t − me

−ktm

k

v0 − mg

k

+

m

k

v0 − mg

k

x(t) = mgk t + m

k (1 − me−

ktm

k

)v0 − mg

k (53)

16 Linear Second Order Equations

• A linear second order equation is an equation that can be written in the form

a2(x)d2y

dx2 + a1(x)

dy

dx + a0(x)y = b(x) (54)

where a0(x), a1(x), 12(x) and b(x) are continuous functions of x on an interval I .

When a0(x), a1(x), 12(x) and b(x) are constant we say that the equation has constant coeffi

cients, otherwise it has variable coefficients.

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• The standard form for linear second order equation,

d2

ydx2

+ p(x) dydx

+ q (x)y = g(x) (55)

where a0(x) ̸= 0 on I , p(x) = a1(x)a2(x) , q (x) = a0(x)

a2(x) and g(x) = b(x)

a2(x)

• g(x) = 0 ⇒ homogeneousg(x) ̸= 0 ⇒ non-homogeneous

16.1 Differential Operator

• Given any function y with a continuous second derivatives on the interval I , theny′′

+ p(x)y′

(x) + q (x)y(x)

generates a new function that we will denote by L[y]. This is,L[y] = y

′′

+ py′

+ qy

• The image of a function y under the operator L is the function L[y].• If we want to evaluate this image function at some point x, we write L[y](x).

Example 1:

Let p(x) = x and q (x) = x − 1Then L[y](x) = y

′′

(x) + xy′

(x) + (x − 1)y(x)

If y1(x) = x3, we get

L[y1](x) = 6x + x(3x2) + (x − 1)x3

= x4 + 2x3 + 6x

So, L maps the function x3 to the function x4 + 2x3 + 6x.

In particular, at x = 2,

Then L[y](x) = 24

+ 2(23

) + 6(2) = 44

• The differential operator L has two very important properties :1.L[y1 + y2] = L[y1] + L[y2]

2.L[cy] = cL[y]

• An operator that satisfies both of the properties for any constant c and any functions y1 and

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y2 in its domain is called a linear operator , otherwise it is called a non-linear operator .

Example 2 :

Show that the operator T defined by

T [y](x) = y′′

(x) + sin(y(x))

where y is any function whose second derivatives is continuous for all real x, is nonlinear.

Solution :

Let y1 = x,

T [cy1](x) = 0 + sin(cx)

= sin(cx)

cT [y1](x) = c[(0) + sin(x)]

= csin(x)

But, in general, sin(cx) ̸= csin(x) for exp : c = 2 and x = π2

sinπ = 0, 2sin(

π2

) = 2

So, second property is violated. Then T is a nonlinear operator.

16.2 Linear Combinations of Solutions

Theorem 4

Let y1 and y2 be solutions to the homogeneous equation

y′′

+ py′

+ qy = 0 (56)

Then any linear combination c1y1 + c2y2 of y1 and y2 where c1 and c2 are constants, is also a

solution to (56).

Proof :

If we let L[y] = y′′

+ py′

+ qy , then L[y1] = 0 and L[y2] = 0, since y1 and y2 are solution to (56).

Using linearity of L as expressed in properties 1 and 2, we have

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L[c1y1 + c2y2] = L[c1y1] + L[c2y2]= c1L[y1] + c2L[y2]

= 0 + 0

= 0

Then c1y1 + c2y2 is a solution to (56).

Example 1:

Given that y1(x) = e2xcos 3x and y2(x) = e2xsin 3x are solutions to the homogeneous equation

y′′ −4y′ + 13y = 0. Find a solution to this equation that satisfies the initial conditions y(0) = 5

and y′

(0) = −2.Solution :

As a consequence of theorem 4, any linear combination

y(x) = c1e2xcos 3x + c2e

2xsin 3x (57)

with c1 and c2 arbitrary constants, will be a solution to y′′

−4y

′

+ 13y = 0.

y′

(x) = c1[2e2xcos 3x − 3e2xsin 3x] + c2[2e2xsin 3x + 3e2xsin 3x] (58)

Substitute (57) and (58) into initial conditions yields c1 = 5, 2c1 + 3c2 = −2 ⇒ c2 = −4So, the solution that satisfies the initial conditions : y(x) = 5e2xcos 3x − 4e2xsin 3x.

16.3 Linear Dependence of Functions

Definition

Two functions y1 and y2 are said to be linearly dependent on an interval I if there exist constants

c1 and c2, not both zero, such thatc1y1(x) + c2y2(x) = 0

for all x in I . If two functions are not linearly dependent, they are said to be linearly independent.

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Example 1:

Determine whether the following pairs of functions y1 and y2 are linearly dependent on (

−5, 5):

(a) y1 = e3x, y2(x) = x + 1

(b) y1(x) = sin 2x, y2(x) = cos xsin x

(c) y1(x) = x, y2(x) = |x|Solution :

(a) A glance at the functions y1(x) = e3x and y2(x) = x +1 indicates that neither is a constant

multiple of the other. Indeed, if a constant c exists such thate3x = c(x + 1) for all x in (−5, 5)

then we arrive at a contradiction by setting x = 0 and x = 1 ;

e0 = c(0 + 1) ⇒ c = 1e3 = c(1 + 1) ⇒ c = e3

2 ̸= 1

So, e3x and x + 1 are linearly independent.

(b) Because y1(x) = sin 2x = 2sin x cos x. So, y1(x) = 2y2(x). Hence y1 and y2 are linearly

dependent on (−5, 5).

(c) y1(x) = x and y2(x) = |x| are identical on the subinterval (0, 5). [In particular, they arelinearly dependent on (0, 5)]. However, neither function is a fixed constant multiple of the

other on the whole interval (−5, 5). [On (0, 5), y1(x) = 1 · y2(x) but on (−5, 0),y1(x) = (−1) · y2(x)]. Thus y1 and y2 are linearly independent on (−5, 5).

16.4 Representation of Solution (Homogeneous Case)

Theorem 5

Let y1 and y2 denote two solutions on (a, b) of y′′

+ p(x)y′

+ q (x)y = 0 (59)

where p and q are continuous on (a, b). Suppose at some point x0 in (a, b) these solutions satisfy

y1(x0)y2′

(x0) − y1′(x0)y2(x0) ̸= 0 (60)

Then every solution of (59) on (a, b) can be expressed in the form

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y(x) = c1y1(x) + c2y2(x)

where c1 and c2 are constants.

Because the expression on the left-hand side of (60) plays an important role in the theory, we

adopt the following terminology.

16.5 Wronskian

Definition : For any two differentiable funstions y1 and y2, the function

W [y1, y2](x) = y1(x)y2′ − y1′(x)y2(x)

=

y1(x) y2(x)y1′(x) y2′(x) (61)

is called the Wronskian of y1 and y2.

16.6 Fundamental Solution Set

A pair of solutions {

y1

, y2}

of y′′

+ py′

+ qy = 0 on (a, b) is called a fundamental solution set if

W [y1, y2](x0) ̸= 0 for all x0 in (a, b).

Example 1:

y1 = ex, y2 = e

−x

W [y1, y2](x) = y1(x)y2′ − y1′(x)y2(x)

= ex e−x

ex

−e−x

= ex(−e−x) − ex(e−x)= −2 ̸= 0

(62)

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17 Procedure for Solving Homogeneous Equation

To determine all solution to y′′

+ py′

+ qy = 0,

(a) Find two solutions y1, y2 that constitue a fundamental solution set

(b) Form the linear combinationy(x) = c1y1(x) + c2y2(x)

where c1 and c2 are arbitrary constants to obtain a general solution.

Example 1:

Given that y1(x) = cos 3x and y2(x) = sin 3x are solutions to

y′′

+ 9y = 0 (63)

on (−∞, ∞), find a general solution to (63).Solution :

First, we verify that {cos 3x, sin 3x} is a fundamental solution set.

W [y1, y2](x) =

cos 3x sin 3x

−3sin 3x 3cos 3x

= (cos 3x)(3cos 3x) − (−3sin 3x)(sin 3x)= 3cos2 3x + 3sin2 3x

= 3[cos2 3x + sin 3x]

= 3(1)

= 3 ̸= 0

Thus, {cos 3x,sin 3x} forms a fundamental solution set, and a general solution to (63) isy(x) = c1cos 3x + c2sin 3x

.

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Theorem 6

If y

1, y

2 are solutions to y

′′

+ p

(x

)y′

+ q

(x

)y

= 0⇓So, y1, y2 are linearly dependent if and only if W [y1, y2] = 0

⇓and it means, y1, y2 are linearly independent if and only if W [y1, y2] ̸= 0

17.1 Fundamental sets, Linear Independence and The Wronskian

If y1, y2 are solutions to y′′

+ p(x)y′

+ q (x)y = 0 on (a, b), then the following statements areequivalent :

(i) y1, y2 is a fundamental solution set on (a, b).

(ii) y1 and y2 are linearly independent on (a, b).

(iii) The Wronskian W [y1, y2](x) is never zero on (a, b).

Example 1:

Show that y1(x) = x−1 and y2(x) = x3 are solutions to

x2y′′ − xy′ − 3y = 0 (64)

on the interval (0, ∞) and give a general solution.Solution :

Substitute yi(x), yi′

(x) and yi′′

(x) i = 1, 2 and into (64) ;

x2(2x−3) − x(−x−2) − 3(x−1)

= 2x−1

+ x−1

− 3x−1

= 0

x2(6x) − x(3x2) − 3(x3)= 6x3 − 3x3 − 3x3= 0

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The solution function x−1 and x3 are linearly independent on (0, ∞) [neither is a constantmultiple of the other on (0,

∞)].

W [y1, y2](x) =

x−1 x3

−x−2 3x2

= 3x1 + x1

= 4x ̸= 0

So, a general solution is : y(x) = c1x−1 + c2x3.

18 Homogeneous Equation With Constant Coeficient

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• ay′′ + by′ + cy = 0, a(̸= 0), b , c

constantsif y1 and y2 are linearly independent, the general solution is : y = c1y1 + c2y2Let say y = erx

ar2erx + brerx + cerx = 0

erx[ar2 + br + c] = 0

Since erx̸= 0, then ar2 + br + c = 0.So, y = erx is a solution to ay

′′

+ by′

+ cy = 0 if and only if r satisfies ar2 + br + c = 0

auxiliary equation.

18.1 Distinct Roots

r1 = −b + √ b2 − 4ac

2a , r1 =

−b − √ b2 − 4ac2a

b2 − 4ac > 0 ⇒ r1, r2 are the distinct real root.So, er1x er2x are the independent solution to ay

′′

+ by′

+ cy = 0.

It can be shown that W [er1x, er2x)

̸= 0]

Then the general solution : y = c1er1x + c2er2x where c1 and c2 are an arbitrary constants.

18.2 Repeated Roots

r1 = −b + √ b2 − 4ac

2a , r1 =

−b − √ b2 − 4ac2a

if b2 − 4ac = 0 ⇒ r1 = r2 = − b2a

then y1 = e− b

2ax

How do you get a second solution which is linearly independent to the first solution?

Let y = v(x)y1

⇒ y′ = v′(x)y1 + v(x)y1′

⇒ y′′ = v′′(x)y1 + v ′(x)y1′ + v ′(x)y1′ + v(x)y1′′

= v′′

(x)y1 + 2v′

(x)y1′

+ v(x)y1′′

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Substitute y, y′

and y′′

intoay

′′

+ by′

+ cy = 0 (65)

ay

′′

+ by

′

+ cy = 0a(v

′′

(x)y1 + 2v′

(x)y1′

+ v(x)y1′′

) + b(v′

(x)y1 + v(x)y1′

) + cv(x)y1 = 0

av′′

(x)y1 + 2av′

(x)y1′

+ av(x)y1′′

+ bv′

(x)y1 + bv(x)y1′

+ cv(x)y1 = 0

(ay1) (iii)

v′′

(x) + (2ay1′

+ by1) (ii)

v′

(x) + (ay1′′

+ by1′

+ cy1) (i)

v(x) = 0

(i) ay1′′

+ by1′

+ cy1 = 0 from (65)

(ii) 2ay′

+ by1 = 0

y1 = e− b2ax ⇒ y1′ = − b

2ae−

b2a

x

Then : 2a

− b

2ae−

b2a

x

+ b

e−

b2a

x

= 0

So, 2ay1′

+ by1 = 0

(iii) ay1v′′

(x) = 0

a ̸= 0 and y1 = e− b2ax̸= 0So, v

′′

(x) = 0

v′

(x) = c1

v(x) = c1x + c2 (66)Substitute (66) into y = v(x)y1, we get :

y = (c1x + c2)y1

= c1xy1 + c2y1

= c1xe− b2ax + c2e−

b2a

x.

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18.3 Complex Roots

r1 = −b + √ b2 − 4ac

2a , r1 =

−b − √ b2 − 4ac2a

if b2 − 4ac


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(c) 4y′′

+ 4y′

+ y = 0

(d) (c) with initial condition; y(2) = 1, y′

(2)

(e) 16z ′′ − 56z ′ + 49z = 0

(f) y′′

+ 2y + 4y = 0

Solution :

(a) 6y′′ − 11y′ + 3y = 06r2 − 11r + 3 = 0

(3r − 1)(2r − 3) = 0r1 =

1

3, r

2 =

3

2 ⇒ 2 distinct real rootSo, e

13

x and e32

x are linearly independent solution.

Hence, the general solution : y(x) = c1e13 x + c2e

32 x.

(b) 2y′′ − 3y′ = 0

2r2 − 3r = 0r(2r − 3) = 0

r1 = 0, r2 = 32

⇒ 2 distinct real root

So, e13

x and e32

x are linearly independent solution.

Hence, the general solution : y(x) = c1e0x + c2e

32

x

= c1 + c2e32

x.

Example 2 :

In the study of a vibrating spring with damping, we are led to an initial value problem of the

form

mx′′

(t) + bx′

(t) + kx(t) = 0; x(0) = x0, x′

(0) = v0 (67)

where m is the mass of the spring system, b is the damping constant, k is the spring constant,

x0 is the initial displacement. v0 is the initial velocity, and x(t) is the displacement from

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equilibrium of the spring system at time t. Determine the equation of motion for this spring

system when m = 36kg, b = 12kgsec−1, k = 37kgsec−2, x0 = 70cm, and v0 = 10cmsec−1.

Also find x(10), the displacement after 10 sec.

Solution :

The equation of motion is given by x(t), the solution of the initial value problem (67) for the

specified values of m, b, k,x0 and v0. That is we seek the solution to

36x′′

+ 12x′

+ 37x = 0; x(0) = 70, x′

(0) = 10 (68)

The auxiliary equation for (68) is : 36r2 + 12r + 37 = 0

which has roots

r1 = −12 ± √ 144 − 4(36)(37)

72 =

−12 ± √ 1 − 3772

= −16 ± i

Hence with α = −16 , β = 1, the displacement x(t) can be expressed in the form

x(t) = c1e− t6 cos t + c2e−

t6 sin t

⇒ x′(t) =−c1

6 + c2

e−

t6 cos t +

−c1 − c2

6

e−

t6 sin t

x(0) = 70 : c1 = 70

x′

(0) = 10 : −c16

+ c2 = 10

⇒ c1 = 70, c2 = 653So, x(t) = 70e−

t6 cos t +

65

3 e−

t6 sin t

and

So, x(10) = 70e−53 cos 10 +

65

3 e−

53 sin 10 ≈ −13.32cm

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19 Higher Order Differential Equation

The method of solving homogeneous linear second order differrential equation and constant co-efficient can be used to solve any order of differential equation. In general, linear homogeneous

differential equation of n order have n independent solutions.

Example 1:

Find a general solution to y′′′

+ y′′ − y′ − 3y = 0.

Solution :

Auxiliary equation : r3 + 3r2

−r

−3 = 0

It can be factorize as : (r − 1)(r + 1)(r − 3) = 0So, r1 = 1, r2 = −1, and r3 = 3Then, the general solution : y(x) = c1e

x + c2e−x + c3e

Example 2 :

Find a general solution to y(4) − y(3) − 3y′′ + 5y′ − 2y = 0.Solution :

Auxiliary equation : r4

−r3

− 3r2

+ 5r

− 2 = 0So, r1 = 1, r2 = 1, r3 = 1, and r4 = −2.Then, the general solution : y(x) = c1e

x + c2xex + c3x

2ex + c4e−2x

Example 3 :

Find a general solution to y′′

+ 2y′

+ 4y = 0.

Solution :

Auxiliary equation : r2 + 2r + 4 = 0

So, r = −2±√ 4−162 = −1 ± i√ 3α = −1, β = √ 3Then, the general solution : y(x) = c1e

−xcos √

3x + c2e−xsin

√ 3x

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20 Superposition and Non-Homogeneous Equations

The differential operator L defined by

L[y](x) = y′′

+ p(x)y′

+ q (x)y (69)

can be viewed as a ”black box” with input the function y(x) and output the function on the

right-hand side of (69).

Suppose the input functions y1(x) and y2(x) yield, respectively, the output functions g1(x) and

g2(x): that isL[y1] = g1(x), L[y2] = g2(x)

Then, since L is alinear operator, an input consisting of the linear combination c1y1(x) + c2y2(x)

produces the output c1g1(x) + c2g2(x) in the same combination.

20.1 Superposition Principle

Let y1 be a solution of the differential equationL[y1] = g1(x)

and let y2 be a solution of

L[y2] = g2(x)where L is a linear differential operator. Then for any constants c1 and c2, the function c1y1 + c2y2

is a solution to the differential equationL[y] = c1g1(x) + c2g2(x)

Example 1:

Given that y1(x) = −x3 − 29 is a solution toy′′

+ 2y′ − 3y = x

and y2(x) = e2x5 is a solution to

y′′

+ 2y′ − 3y = e2x

Find a solution to y′′

+ 2y′ − 3y = 4x − 5e2x

Solution :

Let L[y] = y′′

+ 2y′ − 3y. We are given that

L[y1] = g1(x) = x and L[y2] = g2(x) = e2x

Since we can express 4x − 5e2x = 4g1(x) − 5g2(x),

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then, by the superposition priciple,

c1y1 + c2y2

= 4

−x

3 − 2

9

− 5

e2x

5

= −4x

3 − 8

9 − e2x

is a solution to : L[y](x) = c1g1(x) + c2g2(x) = 4g1(x) − 5g2(x)

20.2 Theorem 7

Let y p(x) be a particular solution to the non-homogeneous equation

y′′

+ p(x)y′

+ q (x)y = g(x) (70)

on the interval (a, b) and let y1(x), y2(x) be linearly independent solutions on (a, b) of the corre-

sponding homogeneous equationy′′

+ p(x)y′

+ q (x)y = 0

Then every solution of (70) on the interval (a, b) can be expressed in the form

y(x) = y p(x) + c1y1(x) + c2y2(x)

⇒ General solution of (70)

20.3 Procedure For Solving Non-Homogeneous Equations

To solve y′′

+ py′

+ qy = g :

(a) Determine a general solution c1y1 + c2y2 of the corresponding homogeneous equation.

(b) Find a particular solution, y p, of the given non-homogeneous equation.

(c) Form the sum of the particular solution and a general to the homogeneous equation; that is,y = y p + c1y1 + c2y2

to obtain a general solution to the given equation.

Example 1:

Given that y p = x2 is a particular solution to

y′′ − y = 2 − x2 (71)

Find a general solution of (71).

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Solution :

Let y′′

−y = 0

⇐ Homogeneous equation

Auxiliary equation : r2 − 1 = 0Then, r = ±1 ⇒ A general solution to the homogeneous equation is c1ex + c2e−x.Combining this particular solution y p = x

2 of the no-homogeneous equation (71), we find that

a general solution isy(x) = x2 + c1e

x + c2e−x

21 Method of Undetermined Coefficients

Method of Undetermined Coefficients is a simple procedure for finding a particular solution to a

non-homogeneous term g(x) is of a special type.

Example 1:

Find a particular solution toL[y](x) = y

′′

+ 3y′

+ 2y = 3x + 1

Solution :

We want to find a function y p(x) such that L[y p](x) is a linear function of x, namely, 3x + 1.

Now what kind of functions y p ”end up” as linear functions after applications of the operator

L Certainly if L[y p](x) linear, so is L[y p](x). So, let’s try the formy p(x) = Ax + B

and attempt to match up L[y p](x) with 3x + 1. SinceL[y p](x) = 0 + 3A + 2(Ax + B) = 2Ax + (3A + 2B)

we need to solve the equation2Ax + (3A + 2B) = 3x + 1

Two polynomials are equal when their corresponding coefficients are equal, so we set

2A = 3

3A + 2B = 1

Solving this system gives A = 32

and B = −74

. Thus the function

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69/70

Example 3 :

Find a particular solution to

L[y](x) = y′′ − y′ − y = sin x

Solution :

We seek a function y p that satisfies L[y p](x) = sin x. Our initial action might be to guess

y p(x) = Asin x, in which case we gety p(x) = −2Asin x − Acos x

However, since the right-hand side of () involves only sin x, this choice of solution would force

A (and hence y p itselft) to be zero. This will not work, since y p(x) = 0 is obviously not a

solution.

So, to compensate for the cos x term, let’s tryy p(x) = Acos x + Bsin x

which we can verify givesL[y p](x) = (−2A − B)cos x + (A − 2B)sin x

So, −2A − B = 0A − 2B = 1

⇒ A = 15 and B = −25 .Hence,

y p(x) = 1

5cos x − 2

5sin x

is a particular solution.

More generally, for an equation of the form

L[y](x) = a cos αx + b sin βx

,

the method of undetermined coefficients suggest that we guess

y p(x) = A cos αx + B sin βx

and solve y p(x) = 15

cos x − 25

sin x for the unknowns A and B.

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Example 4:

Find a particular solution to

L[y](x) = y′′

+ 4y′

= 3cos 2x

Solution :

We first guess y p(x) = Acos 2x + Bsin 2x

L[y p](x) = (4A − 4A) =0

cos 2x + (4B − 4B) =0

sin 2x = 3cos 2x

,

It means that y p(x) = Acos 2x + Bsin 2x is not a particular solution for y′′

+ 4y′

= 3cos 2x

since the left hand side is equal to zero. Let’s find the solution to the corresponding homoge

neous equation, y′′ + 4y′ = 0

Auxiliary equation : r4 + 4 = 0

r = ±2i

Then a general solution to the homogneous equation isc1e

0cos 2x + c2e0sin 2x = c1cos 2x + c2sin 2x

Now we assumey p(x) = Axcos 2x + Bxsin 2x

Then L[y p](x) = −4Asin 2x + 4B cos 2x = 3cos 2x,

where A = 0 and B = 34

. So, we find that a particular solution is

y p(x) = 3

4xsin 2x

.

In the preceding example, the trial choice for y p did not work because it was a solution to

the corresponding homogeneous equation. However, when we replaced y p by the new function

y p = xy p, we were able to find a particular solution to the non-homogeneous equation. Thisprocedure for ”repairing” the method of undetermined coefficients is generalized as follows.

In any term in the trial expression for y p is a solution to the corresponding homogeneous

equation, then yp by xyp. If the latter is also a solution, try x2yp, then x

3yp, etc., ultimately

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