math 54 lecture 5 - differential equations

DE Solution Ortho Trajectories Exponential Growth/Decay

Differential Equations and ApplicationsMathematics 54–Elementary Analysis 2

Institute of MathematicsUniversity of the Philippines-Diliman

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Introduction

Definition.

A differential equation is an equation involving a function of onevariable and its derivatives.

Examples.

1dy

dx= 6−3x2

2du

dy= e−yy3

3d2u

dx2 = tanx sec2 x

4 v′′ = 1

x2 +5

Definitions.1 The order of a differential equation is the highest order of the

derivative in the equation. ((1)–(2) are of order 1, (3)–(4) are oforder 2.)

2 A solution of a differential equation is a function y = f (x) thatsatisfies the equation.

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Differential Equations

Consider the differential equation

dy

dx= tan2 3x

y2 −1.

Notice that the above equation can be written as(y2 −1

)dy = tan2 3x dx

Definition

A differential equation that can be expressed in the formf (x)dx = g(y)dy is said to be separable.

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General Solution of a Differential Equation

Example.

1 Find a solution ofdy

dx= cos3x

sin2y.

Solution: We have

sin2y dy = cos3x dx

⇒∫

sin2y dy =∫

cos3x dx

⇒ −1

2cos2y+C1 = 1

3sin3x+C2

.

Combining C1 and C2, the solution has the form

1

2cos2y+ 1

3sin3x = C , C ∈R

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General Solution of a Differential Equation

Example.

2 Solvedu

dx= x

p1−u2

up

2x2 +1.

Solution: The given equation is separable. Indeed,

up1−u2

du = xp2x2 +1

dx

⇒∫

up1−u2

du =∫

xp2x2 +1

dx

⇒−√

1−u2 +C1 = 1

2

√2x2 +1+C2

⇒ C = 1

2

√2x2 +1+

√1−u2.

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Particular Solution of a Differential Equation

Example.

3 Find the equation of the curve that passes through (0,1) andwhose slope at any point

(x,y

)is given by y cosx

1+y2 .

Solution: The equation of the curve satisfiesdy

dx= y cosx

1+y2 .

⇒ 1+y2

ydy = cosx dx

⇒∫ (

1

y+y

)dy =

∫cosx dx

⇒ ln∣∣y∣∣+ 1

2y2 = sinx+C

But (0,1) is a point on the above curve. Thus, we have

ln |1|+ 1

2(1)2 = sin0+C =⇒ C = 1

2.

The equation of the curve is ln |y|+ 1

2y2 = sinx+ 1

2.

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Orthogonal Trajectories

Definition.

An orthogonal trajectory to a set of curves is a curve that intersectseach curve, in the given set, orthogonally (perperdicularly). That is,the tangent lines at the points of intersection are perpendicular.

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Example.

1 Consider the set of circles with center at the origin x2 +y2 = C.

Some orthogonal trajectoriesare:

x−axis,

y−axis,

y = kx, k ∈R

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Computing Equations of Orthogonal Trajectories

Example.

Consider the set of parabolas with vertex at the origin y = kx2.

Note: if each curve in the set hasslope m,then each curve in the set oforthogonal trajectories has

slopemOT =− 1

m.

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Orthogonal Trajectories of y = kx2

Each curve in the set has slope

m = dy

dx= 2kx

Since this is true for all k, andy

x2 = k =⇒ m = 2( y

x2

)x = 2y

xHence, an orthogonal trajectory must have slope mOT =− x

2y.

That is, an orthogonal trajectory must satisfy the differentialequation

dy

dx=− x

2y.

Solving, we obtain y2 =−1

2x2 +C, where C ∈R.

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Example.

set of curves : {y = kx2 : k ∈R}set of orthogonal trajectories : {y2 + 1

2 x2 = C : C ∈R}

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Orthogonal Trajectories of y = x2 +C

Example.

Consider the set of parabolas y = x2+C, C ∈R, whose vertices are onthe y-axis.

Each curve in the set has slope m = dy

dx= 2x.

Hence, each orthogonal trajectory must have slope

mOT =− 1

m=− 1

2x.

That is, each orthogonal trajectory must satisfy the differentialequation

dy

dx=− 1

2x.

Thus, the solution is y =−1

2ln |x|+K , where K ∈R.

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Example.

set of curves : {y = x2 +C : C ∈R}set of orthogonal trajectories : {y =−1

2 ln |x|+K : K ∈R}

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Orthogonal Trajectories

Example.

Consider the set of curves siny = ex +C.

Computing for the slope m = dy

dxof each curve above :

d

dx

(siny

) = d

dx

(ex +C

)⇒ cosy

dy

dx= ex

⇒ dy

dx= ex secy

Thus, each orthogonal trajectory must satisfy the differentialequation,

dy

dx=− 1

ex secy.

Solving, we obtain ln∣∣secy+ tany

∣∣= e−x +K .14 / 22


Mathematical Modelling

Exponential Growth and Decay

A population is said to be growing (or decaying) exponentially if itsinstantaneous rate of increase (or decrease) at any given time isproportional to the population at that time. (e.g., bacterial cultureand radioactive decay)

Let P = P(t) be the population at any time t.

Hence,dP

dtis the rate of change at any time t. Thus,

dP

dt∝ P

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Exponential Growth and Decay

dP

dt∝ P =⇒ dP

dt= kP, k : constant

1

PdP = k dt

lnP = kt +C

P (t) = ekt+C

P (t) = ekteC

Let t = 0 =⇒ P (0) = eC .

Equation of the Model

P = P(t) = P0ekt

where P0 indicates initial population;if k > 0, the model is for exponential growth;

if k < 0, the model is for exponential decay.

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P (t) = P (0)ekt

Example 1.

There are 1000 bacteria at a time and the number doubles after 30minutes. If the population is proportional to its rate of change,(a) find the population of the culture after t hours.(b) How many will be present in 2 hrs?(c) How long will it take before 64000 is present?

(a) The initial population P0 = 1000. Next goal is to find k.

P(1

2

)= 2000 = 1000e(

12 k

)=⇒ e

(12 k

)= 2 =⇒ ek = 4.

Therefore, P (t) = 1000(4)t .

(b) When t = 2, P(2) = 16000.

(c) We find t such that 64000 = 1000(4)t . Hence t = 3 hrs.

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P (t) = P (0)ekt

Example 2.

A bacterial culture triples its population after an hour. If 81M werepresent at the end of 4hrs, how many were present initially.

We must find P0 = P (0). Note that P (1) = 3P0 = P0ek·1 ⇒ ek = 3.Hence we have

P(t) = P0(3)t

81M = P (4) = P034

Hence, P0 = 1M .

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P (t) = P (0)ekt

"Radioactive substances decay by spontaneous emission of radiation. It’s

been experimentally found that the rate of decay is proportional to the

amount of the substance at any given time".

The half-life of a substance is the time it takes for half of theamount to decay.

Example 3.

Element X has a half life of 3 yrs. If 64g of it is present at a time, howmuch of it will be present after 16 yrs.

Half-life of 3yrs ⇒ P (3) = 12 P0 = P0e3k ⇒ e3k = 1

2 ⇐= ek = (12

)1/3.

Find P (16):

P (16) = P0e16k = 64(ek

)16 = 64(1

2

) 163 = 64

3p

216= 64

32 3p

2= 3p

4 g.

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P (t) = P (0)ekt

Example 4.

30% of a radioactive substance disappears in 15yrs. Find its half-life.

Find t such that P (t) = 12 P0: P (15) = (0.70)P0.

P (15) = P0e15k ⇒ P0e15k = 0.7P0

⇒ e15k = 0.7

⇒ ek = (0.7)1

15 .

We now want to know at t is

P (t) = 12 P0 =⇒ 1

2 P0 = P0ekt

=⇒ 12 = (

ek)t = (0.7)

t15

=⇒ t = −15ln2

ln0.7.

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Exercises

1 Solve the following differential equations.

1dy

dx= xex

y√

1+y2

2dy

dx= ey sin2 x

y secx

3dy

dx= y cosx

1+y2 , y (0) = 1

4 xy′+y = y2, y (1) =−1

2 Do as indicated.1 Find the equation of the curve that passes through the point

(0,1) and whose slope at(x,y

)is xy.

2 Find the function f such that f ′ (x) = f (x)(1− f (x)

)and f (0) = 1

2.

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Exercises

3 Find the orthogonal trajectories of the family of curves.1 x2 +2y2 = K2 y2 = Kx3

3 y = x

1+Kx4 Do as indicated.

1 A bacteria culture starts with 500 bacteria. After 3 hrs there are8000 bacteria. Find: (i) the initial population; (ii) the formulafor the population after t hours; (iii) the number of bacteriaafter 5 hours; (iv) the rate of growth after 5 hours; (v) the timewhen population reaches 200000.

2 Bismuth-210 has a half-life of 5 days. At start, there is an 800mgsample. Find: (i) a formula for the mass remaining after t days;(ii) mass remaining after 30 days; (iii) time when the remainingamount is 1mg.

3 After 3 days a sample of Radon-222 decays to 58% of its originalamount. Find: (i) the half-life of Radon-222; (ii) the time itwould take for the sample to decay 10% of its amount.

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math 54 lecture 5 - differential equations

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