# DIFFERENTIAL EQUATION

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UNAWEZA JIPATIA NOTES ZETU KWA KUCHANGIA KIASI KIDOGO KABISA:PIGA SIMU:0787237719

A differential equation is a relationship between the rates of different variables (i.e an independent variables x, a dependent variable, y, and one or more differential coefficient of y with respect to x).

Eg         + y sin x = 0

xy + y  +  = 0

+  – 4y = 0

The order of a differential equation

The order of a differential equation is given by the highest derivative involved in the equation.

– y2 = 0 is an eqn of the 1st order

xy – y2 sin x = 0 is an eqn of the 2nd order

xy – y + = 0 is an eqn of the 3rd order

The degree of a differential equation

Eg  + y sin x = 0 the degree is 1b

Linear differential equation (L.D.E)

A linear differential equation should look in the form

+  +  + ………   + pn y = Q

Where P1, P2, P3 …Pn, Q are functions of x, or constant n = 1, 2, 3, 4…….

Examples

i)  +  + 6y = 0

ii)  +  +  =

iii)   = 4

Examples of non L.D.E

+  + 6y = 0

+  + 6y = 0

Note

A D. E will be linear if

•          Variable y and its derivatives occur at the first degree only
•          No product of y and its derivatives
•         No transcendental function of y or x

Exercise

Show which of the following differential equations is the L.D.E or N.L.D.E

i)  = y……………linear

ii) (x2 – 1)  = y……linear

iii)  + y2 + 4 = 0…………..non – linear

iv)  = ……………non – linear

v)  = 1 + xy………linear

vi) (x2 – y2)  = 2xy……Non – linear

Formulation of a differential

Differential equation may be formed when arbitrary constant are eliminated from a given function.

Examples 1

Y = A sin x + B cos x, where A and B are two arbitrary constants

Solution

= A cos x – B sin x

= -A sin x – B cos x

his is identical to the original eqn with opposite signs

= -1(A cos x + B sin x)

= -y

Form a DE whose solution is the form y = Ae2x + Be-3X  Where A and B are constant.

Example 2

Form a different equation from the function y = x +

Solution
y = x +  = x + Ax-1

= 1 – Ax-2 = 1 –

From the given equation

Y = x +

yx = x +

x(y – x) = A

= 1- x (y – x) x-2

= 1 –

= x2 – x (y – x)

= x – (y – x)

= 2x – y

Example 3

Form the DE for y = Ax2 + Bx

Y = Ax2 + Bx…… (i)

= 2Ax + B

– 2Ax = B……. (ii)

= 2A

Substituting both (ii) and (iii) into (i) gives

– 2x + 2y = 0

Solution to a DE

This involves finding the function for which the equation is true (i.e manipulating the eqn so as to eliminate all the differential coefficients and have a relationship between x and y)

E.g. verify that the function

i) Y = 3e2x is a soln of DE =  – 2y = 0 for all x

ii) y (x) = sin x – cos x + 1 is a soln of eqn  + y = 1 for all value of x

Solution

Substitute equation iv into iii.

= 2y
– 2y = 0

Direct integration
Direct integration is used to solve equation which is arranged in the form  = f (x)

Examples

1.  Solves   = 3x2 – 6x + 5

Solution

= 3x2 – 6x + 5

Then  =

= x3 – 3x2 + 5x + c

Y = x3 – 3x2 + 5x + c

2.  Solve  = 5x3 + 4

Solution

Rearranging in the form  = f (x)

= 5x2 +

Then   =  dx

Y =  + 4ln x + c

Y =  + 4ln x + c

The solution is a called general solution since it consist a constant C (i.e. unknown constant)

3. Find the solution of the eqn ex  = 4 given that y = 3

When x = 0

Soln: rearrange the given equation

=  = 4e-x + 7

This is called a particular soln since it contains a non – unknown       variable.

First order D.E

Separating the variables

This is a method used to solve the D.E when is in the form  = f (x, y)

The variables y on the right hand-side prevents solving by direct integration

Example

1. Solve   =

Solution   =

= (y + 1)  = 2x

Integrating both sides with respect to x

2. Solve   =

Solution

=

=

dy =  dx

3.   By separating variables solve the differential equation (xy + x) dx =(x2y2 + y2 + x2 + 1) dy

Solution

Given Equation;

(xy + x) dx = (x2y2 + y2 + x2 + 1) dy

X (y + 1) dx = (y2 (x2 + 1) + (x2 +1)) dy

X (y + 1) dx = (x2 + 1) (y2 + 1)  dy

dx =   dy

Integrating both sides

dx =   dy

R.H.S == y-1+

dx =) dy

½ ln (x2 + 1) =  – y + 2 ln (y + 1) + C

First order homogenous D.E

In first order H.D.E all the terms are of the same dimension

 Term X x2 1/x xn 3 y yn x2/y x2y Dimension 1 2 -1 n O 1 N 1 3 dy/dx dy2/dx2 xdy/dx yx 0 1 2

O shows there is no effect on the term

Which of the following equation are first order homogeneous

a) x2  = y2

b) xy  = x2 + y2

c) x2  = 1 + xy

d) (x2  =

e) (x2 – y2)  = 2xy

f) (1 + y2)  = x

Solution of 1st order Homogeneous  differential equation

1st order homogeneous d.e can be written in the form  = Q

Where both P and Q are function of and have same dimension.

Suppose p and Q hence the dimension

Then divide by xn and use the substitution

Y = vx =      v =

i.e

Example 1.

Solve  =

Solution

Since all terms are of degree 2, i.e the equation is homogeneous

Let y = vx

= v +

And

=

=  ()

=

V +  =

=  – v

=

=

=

=  dx

= lnx + c

But v =

2 = ln x + c

= lnx + c

Y2 = 2x2 (lnx + c)

Example 2

Solve xy = x2 + y2

Solution

All terms are of the order 2

xy  = x2 + y2……

Let y = vx

= v +

From the equation…

=  +

But y = vx

=  =  +

= x2

=

V +  =

=

=

=

=

= lnx + A

But v =

2 = lnx + A

Y2 = 2x2 (lnx + A)

Example 3

Solve the following

(x2 + xy)  = xy – y2

Solution

Let y = vx

= v +

And

=

=

v +  =

= -v

=

=

=

[( dv +  dv)] =

-½ ( + ln v) = lnx + c

= ln (x2 V) + 2c

But v =

= ln (x2) + 2c

– 2c = ln (x2.)

Let ln A = -2c

+ ln A = ln xy

Xy = Aex/y

EXERCISE

Solve the differential equation

1.  x2  = y (x + y)

2. (x2 + y2)  = xy

First order exact differential equation

We know that

Now consider

+ y = ex

Integrating both side w.r.t.x

= xy = e2 + c

Y =  (ex + c)

Y =  (ex + C)

2.  Find the general solution of the following exact differential equation.

i)exy + e= 2

ii) cos x – ysin x = x2

Solution

i) Given

(exy) = 2

Integrating both sides with respect to x

(exy) dx =

ii) Given

cosx  – y sin x = x2

(y cos x) = x2

Integrating both side with respect to x

=  (exy) dx =

Y cos x =  + c

3ycosx = x3 + 3C

3ycosx = x3 + 4

3.        + lny = x + 1

Soln

Given   + lny = x + 1

Integrating both sides w.r.t x

(x ln y) dx =

Xln y = ½ x2 + x + C

2xlny + x2 + 2x + 2c

2x lny + x2 + x + A

Integrating factors

Consider first order differential equation of the form

+ py =Q

Where p and Q are functions of x

Multiplying by integrating factor F both sides will make an exact equation

i.e F  + Fpy + FQ…….i

=  +

+ …..ii

Comparing (i) (ii)

+ Fpy =  +

Fpy =

= Fp

dF =  by separating of variables

F =  is the required integrating factor

Examples

1. Solve  + y = x3

Solution

+ y = x3

+  = x2

Compare with  + py = Q

P =

Then  =  dx

= ln x

If F = eln x

F       = eln x

(xy) = x2 x

(xy) =

Xy =  x4 + C

2. Solve

(x + 1)  + y + (x + 1)2

Soln

(x + 1)  + y + (x + 1)2

=  +  = x + 1

=  + , y = x + 1

P =

F = esp

= eln (x + 1)

= x + 1

Y (x + 1) =

=

= x3 + x2 + x + c

=  + c

Y =  +

3.      Solve (1 – x2)  – xy = 1

Solution   –  y =

P =

=  dx

= ½ ln (1 – )

F = e ln (1 –  ½

= (1 – x2) ½

Y  =   dx

=  dx

Y =

4.      tanx  + y = ex tan x

Solution

+ , y = ex

P =  = cot x

=

= ln

F = eln

=

Y =  dx

Integrating by parts the R.H.S

=  = ex –  dx

= ex sin x – (ex cos x +  dx

= ex – ex –  dx

2 sin x dx = e x ( –)

=  ()

Y sin x =  () + c

Bernoulli’s equation

This is a first order D.E of the form

+ p (x) y = Q (x) y n

p (x) and Q (x) are functions of x or constant

Steps

+ p (x) y = Q (x) yn……i

Divide both sides by yn gives

Y-n   + p (x) y1 – n = Q (x)…… (ii)

Let z = y1-n

(1 – n) y-n

Multiplying (ii) by (1 – n) both sides gives

(1 – n) y-n   + (1 – n) p (x) y (1 – n) = (1 – n) Q (x)

+ (1 – n) p (x) y (1 – n) = (1 – n) Q (x)

+ p1 (x) y (1 – n) = Q1 (x)

p1 (x) and Q1 (x) are functions of x or constant

But z = y (1 -n)

=  + p1 (x) z = Q1 (x)……..ii

iii) is linear the use of integrating  factor  can be used

Example 1

Solve  +  = xy2

Solution

Dividing both sides by  gives

Let Z =

Multiply (ii) by -1 gives

But Z =

=

=

=

Then Z.F=

= -1

=

But Z =

Example 2

Solve

Solution

1st Expressing the equation in the form

Let Z =

But Z =

=

=

Then

But Z =

Example 3

Solve

Solution

Expressing the equation in the form

Then dividing by

let  Z =

Multiply (ii) by -2 gives

But Z =

Then,

But Z =

Exercise

Solve the following first order D.es

1.

2.

Second order Differential Equations

Second order differential equation is of the form of;

Where a, b, c are constant coefficients and is a given function of x

If   the equation is homogeneous otherwise it is a non-homogeneous

Which of the following are linear H.D.Es

1.

2.

3.

4.

5.

Characteristic (Auxiliary) Equation for H.DE

Consider a linear non-homogeneous 2nd order D.E

Let y = u and y=v be two solution of the equation

Where u and v a functions of x

Then,

and

Becomes

If y = u and y = v are the solutions of the equation

Suppose a = 0

i.e.

(separable)

(where is constant)

Take m for –k

–         is the solution of dy +ky

Also will be the solution of the equation

if it satisfies the equation

If

The values will be  and

If   are two solution

Note that:
If the Auxiliary equation has

(i)   Two real roots

(ii)    Equal roots

The solution is

(iii)  2 Complex roots to the auxiliary equation

The solution is

Examples

Solve the following 2nd order Des

1.

Solution

Auxiliary equation

2.

Solution

The auxiliary equation is

3.

Solution

Auxiliary equation is

In this case P=-2 and q =

Exercise

Solve the following

1.

2.

3.

Non homogeneous 2nd order D.E

Consider the equation

If

 General solution = complementary function +particular integrate

The general solution of (i) is given by

Note that

is called complementary function matas R.H.S Zero

Y = f (x) is called particular intergral makes R.H.S 0

Consider the R.H.S function

i.e if          assume

Examples

1.     Solve

Solution

C.F solve L.H.S = 0

Assume

Substituting in the given equation gives

2.     Solve

Solution

(i)   C.F:

(ii)

Substituting into the given equation

Collecting like terms

Comparing L.H.S to R.H.s

General solution

3.     Solve

Given that

Solution

C.F:

I: Assume

+2

Substitute into (i)

But y = 1,

1= A + B + 1

A + B = 0

A-2B+3

………………………….(i)

B = 3

Substituting into (ii) gives

A + 3 = 0

A = -3

4.     Solve

Solution

C.F:

Let y =

5

2

2C=1

Note that: if P.I is contained in the C.F multiply the assumed P.I by x and go on

Example

Solve

Solution

C.F:

P.I: Assume y =

+B

Exercise

Solve the following

(i)

2nd order equations which are reducible to 1st order

Consider 2nd order equation which can not be written in the form

i.e
Such equation will be solved by the substitution of: –

Where A and B are constants
Example
Solve
Solution
Let

dx

Example

Solve

Solution

Let p =

Example

Solution

Let

In p= ln Ay where c= ln A
P = Ay

i.e.

Note that:

– If  P.I is contained in C.F, and C.F is of real  roots Assume  if

– If  C.F is a distinct root, but one root is the  same as that on P.I, assume if

– If  , assume  independently followed by  then add to obtained a P.I

Applications of differential Equations

1.    i.  Population growth

The rate of population growth id differently proportional to the number of inhabitants present at specific time.

If N is the number of inhabitants at specific time.

When

i.e.

growth equation

If the population of a certain place  doubles in 50 years. After how many years will the population treble and the assumption that the rate of increase is proportional to the number of inhabitant in the place.

Solution

Let N be the number of inhabitants at time t.

When

………………………..(i)

When t = 50 years,   N=2

=

K =

Substituting into (i)

ln N = kt +ln No

Example

The population of Kenya is known to increase to a rate proportional to the number of people living there at any given time. In 1986 the population was 1.1 times that of 1984 and 1987 the population was 18,000,000. What was the population in 1984?

Solution

Let N be the number of people at time, t

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