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DIFFERENTIAL EQUATIONS Introduction

```MODULE 1: DIFFERENTIAL EQUATIONS
Invention of Differential Equations:
-In mathematics, the history of DE traces the development of “DE” from calculus,
which itself was independently invented by English physicist Isaac Newton and
German mathematician Gottfried Leibniz.
-The history of the subject of differential equations, in concise form, from a synopsis
of the recent article “The History of Differential Equations, 1670-1950”
- “Differential Equations” began with Leibniz, the Bernoulli brothers, and others
from the 1680’s, not long after Newton’s fluxional equations’ in the 1670’s.
Differential equations
-
is an equation involving an unknown function and its derivatives.
is an equation which contains derivatives or differential is called a differential
equation.
A differential equation is an ordinary differential equation if the unknown function
depends on only one independent variable. If the unknown function depends on two
or more independent variables, the differential equation is a partial differential
equation.
NOTE: Equations 1.1 through 1.4 are examples of ordinary differential equations,
since the unknown function y depends solely on the variable x. Equation 1.5 is a
partial differential equation, since y depends on both the independent variables t and
x.
TYPES OF ORDINARY DIFFERENTIAL EQUATIONS
 First Order ODE
-First Order Linear ODE
-Exact Equation
-Non-Linear First order
-Separable Equation
-Bernoulli Differential Equation
 Second Order ODE
-Linear Second Order ODE
-Homogeneous Secondary Order ODE
-Initial and Boundary Value Problems
-Non-Linear Second Order ODE
-Non-Homogeneous Second Order ODE
 Higher Order ODE
-Linear nth Order ODE
-Homogeneous Equation
-Non-Homogeneous Equation
Order and Degree of a Differential Equation
a) Order of a Differential equation
-this is the order of the highest derivative that appears in the D.E.
Example:



𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦
− 5𝑥 𝑑𝑥 = 2𝑒 𝑥 + 1
2
𝑥𝑠𝑖𝑛𝑦𝑑𝑥 + 𝑥 𝑐𝑜𝑠𝑦𝑑𝑦 = 0
𝑑3 𝑦
𝑑𝑡 3
𝑑𝑦
+ 𝑡 𝑑𝑡 = 𝑦 2 𝑡 − 3
→ 𝑠𝑒𝑐𝑜𝑛𝑑 𝑜𝑟𝑑𝑒𝑟
→ 𝑓𝑖𝑟𝑠𝑡 𝑜𝑟𝑑𝑒𝑟
→ 𝑡ℎ𝑖𝑟𝑑 𝑜𝑟𝑑𝑒𝑟
b) Degree of a Differential Equation
If a differential equation can be written as a polynomial in the unknown
function and its derivatives, then its degree is the power to which the highest
order derivative is raised.



𝑑𝑦 3
5𝑥 [𝑑𝑥 ] − 3𝑦 = 𝑒 2𝑥
𝑑2 𝑦
𝑑𝑥 2
𝑑𝑦 4
− 3𝑥 [𝑑𝑥 ] + 2𝑥 2 + 5
𝑑3 𝑠
2
𝑑𝑠 5
[𝑑𝑡 3 ] + [𝑑𝑡 ] = 3𝑡 − 4
→ 𝑡ℎ𝑖𝑟𝑑 𝑑𝑒𝑔𝑟𝑒𝑒
→ 𝑓𝑖𝑟𝑠𝑡 𝑑𝑒𝑔𝑟𝑒𝑒
→ 𝑠𝑒𝑐𝑜𝑛𝑑 𝑑𝑒𝑔𝑟𝑒𝑒
Linear and Non-Linear Differentials
An ordinary differential equation of order “n” is called linear if it has the form:
𝒂𝟎 (𝒙)
𝒅𝒏 𝒚
𝒅𝒏−𝟏 𝒚
𝒅𝒚
(𝒙)
+
𝒂
+ ⋯ + 𝒂𝒏−𝟏 (𝒙)
+ 𝒂𝒏 (𝒙)𝒚 = 𝒇(𝒙)
𝟏
𝒏
𝒏−𝟏
𝒅𝒙
𝒅𝒙
𝒅𝒙
Otherwise, if it can’t then, it is non-linear.
Examples:
a)
𝑑𝑦
+ 𝑃(𝑥)𝑦 = 𝑄(𝑥)
→ 𝑙𝑖𝑛𝑒𝑎𝑟
𝑑𝑥
𝑑𝑦
b) 𝑦 𝑑𝑥 − 2𝑥 2 𝑦 = 𝑥 3 − 1
𝑑3 𝑥
→ 𝑛𝑜𝑛 − 𝑙𝑖𝑛𝑒𝑎𝑟
𝑑𝑥
c) 5𝑡 𝑑𝑡 3 + 2𝑡 2 𝑑𝑡 − 𝑥𝑠𝑖𝑛 𝑡 = 0
𝑑2 𝑧
→ 𝑙𝑖𝑛𝑒𝑎𝑟
𝑑𝑧
d) 𝑥 𝑑𝑥 2 + 4𝑥 3 𝑑𝑥 − √𝑧 cos 𝑥 = 5𝑥 2 − 3 → 𝑛𝑜𝑛 − 𝑙𝑖𝑛𝑒𝑎𝑟
Assignment # 1
1.
2.
State the order, degree, and linearity of the following differential equations.
𝑑5 𝑦
𝑑𝑦 4
2𝑥 3 5 + cos 𝑥 ( ) = 3𝑥 − 4
𝑑𝑥
𝑑𝑥
2 ′′′
′
𝑡 𝑦 − 3𝑡𝑦 = 1 + sin 𝑡
3. 𝑥 2 𝑦
4.
𝑑6 𝑠
𝑑𝑡 4
𝑑2 𝑦
𝑑𝑥 4
+ (𝑥
𝑑𝑦
𝑑𝑥
− 2𝑦)2
− 2𝑠𝑡 3 = 𝑡 + 7
Elimination of Arbitrary Constant:
This is the process of eliminating the arbitrary constant present in a given
equation to obtain the required differential equation.
The number of arbitrary constants present suggests how many times
differentiation should take place.
Sample 1: Eliminate the arbitrary constants
a) 𝑦 = 3𝐴𝑥 3 − 2𝐵𝑥 2 + 4𝐶𝑥 − 𝐷
Solution: By successive differentiation
𝑦 ′ = 9𝐴𝑥 2 − 4𝐵𝑥 + 4
𝑦 ′′ = 18𝐴𝑥 − 4𝐵
𝑦 ′′′ = 18𝐴
𝒚𝒊𝒗 = 𝟎
b) 𝑥 2 𝑦 3 + 𝑥 3 𝑦 5 = 𝐶
Solution: By Implicit Differentiation
𝑑𝑦
𝑑𝑦
(2𝑥𝑦 3 + 3𝑥 2 𝑦 2 ) + (3𝑥 2 𝑦 5 + 5𝑥 3 𝑦 4 ) = 0
𝑑𝑥
𝑑𝑥
(2𝑥𝑦 3 𝑑𝑥 + 3𝑥 2 𝑦 2 𝑑𝑦) + (3𝑥 2 𝑦 5 𝑑𝑥 + 5𝑥 3 𝑦 4 𝑑𝑦) = 0 **Dividing both sides by
𝑥𝑦 2 results to,
(2𝑦𝑑𝑥 + 3𝑥𝑑𝑦) + (3𝑥𝑦 3 𝑑𝑥 + 5𝑥 2 𝑦 2 𝑑𝑦) = 0
(𝟐𝒚 + 𝟑𝒙𝒚𝟑 )𝒅𝒙 + (𝟑𝒙 + 𝟓𝒙𝟐 𝒚𝟐 )𝒅𝒚 = 𝟎
c) 𝑦 = 𝐶1 𝑒 2𝑥 + 𝐶2 𝑒 3𝑥 → (1)
Solution:
By successive differentiation
′
𝑦 = 2𝐶1 𝑒 2𝑥 + 3𝐶2 𝑒 3𝑥 → (2)
𝑦′′ = 4𝐶1 𝑒 2𝑥 + 9𝐶2 𝑒 3𝑥 → (3)
*Eliminate 𝑪𝟏 𝒆𝟐𝒙 in (1) and (2);
2 (𝑦 = 𝐶1 𝑒 2𝑥 + 𝐶2 𝑒 3𝑥 )
−
𝑦 ′ = 2𝐶1 𝑒 2𝑥 + 3𝐶2 𝑒 3𝑥
𝟐𝒚 − 𝒚′ = − 𝑪𝟐 𝒆𝟑𝒙
→ (𝟒)
*Eliminate 𝑪𝟏 𝒆𝟐𝒙 in (1) and (3);
4 (𝑦 = 𝐶1 𝑒 2𝑥 + 𝐶2 𝑒 3𝑥 )
−
𝑦′′ = 4𝐶1 𝑒 2𝑥 + 9𝐶2 𝑒 3𝑥
𝟒𝒚 − 𝒚′′ = −𝟓𝑪𝟐 𝒆𝟑𝒙 → (𝟓)
*Eliminate 𝑪𝟐 𝒆𝟑𝒙 in (4) and (5);
−
10𝑦 − 5𝑦 ′ = −5 𝐶2 𝑒 3𝑥
4𝑦 − 𝑦 ′′ = −5𝐶2 𝑒 3𝑥
𝟔𝒚 − 𝟓𝒚′ + 𝒚′′ = 𝟎
Activity #1 : Eliminate the arbitrary constant:
1.
2.
3.
4.
5.
𝑥 3 − 3𝑥 2 𝑦 = 𝑐
𝑦 sin 𝑥 − 𝑥𝑦 2 = 𝑐
𝑦 = 𝑐𝑥 + 𝑐 2 + 1
𝑦 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 −𝑥
𝑦 = 𝐶1 𝑒 2𝑥 + 𝐶2 𝑥𝑒 2𝑥
Solution of a Differential Equation
-
The solution of differential equation is a relation among the variables which
reduces the equation to an algebraic identity.
General and Particular Solution
a) General Solution -a solution which contains a number of independent
constant equal to the order of the differential equation.
b) Particular Solution – a solution which can be obtained from the general
solution by solving the specific value of the arbitrary constant present.
Differential Equation of First Order and First Degree
-
A differential equation of first order and first degree may be written in the form:
𝑴 (𝒙, 𝒚)𝒅𝒙 + 𝑵(𝒙, 𝒚)𝒅𝒚 = 𝟎
Where M and N are both functions of x and y.
Variable-Separable Differential Equations
-if by algebraic process 𝑴 (𝒙, 𝒚)𝒅𝒙 + 𝑵(𝒙, 𝒚)𝒅𝒚 = 𝟎 may be written in the form
𝑴𝟏 (𝒙)𝒅𝒙 + 𝑵𝟏 (𝒚)𝒅𝒚 = 𝟎 where 𝑴𝟏 𝒂𝒏𝒅 𝑵𝟏 are functions of one variable as
indicated, then the variable have been separated. Hence the general solution can be
obtained by integration:
∫ 𝑴𝟏 (𝒙)𝒅𝒙 + 𝑵𝟏 (𝒚)𝒅𝒚 = 𝑪
Sample Problem: Solve the following differential equations.
a) 𝑠𝑖𝑛𝑥 𝑠𝑖𝑛𝑦 𝑑𝑥 + 𝑐𝑜𝑠𝑥 𝑐𝑜𝑠𝑦 𝑑𝑦 = 0
Solution:
1
*Multiplying both sides by sin 𝑦 sin 𝑥 results to
sin 𝑥
cos 𝑦
𝑑𝑥 +
𝑑𝑦 = 0
cos 𝑥
sin 𝑦
∫ tan 𝑥 𝑑𝑥 + ∫ cot 𝑦 𝑑𝑦 = 𝐶
− ln cos 𝑥 + ln sin 𝑦 = 𝐶
sin 𝑦
𝑙𝑛
=𝐶
cos 𝑥
Take the inverse logarithm of the both sides:
sin 𝑦
𝑒 𝑙𝑛cos 𝑥 = 𝑒 𝑐
sin 𝑦
= 𝐶1 ; 𝐶1 = 𝑒 𝑐
cos 𝑥
sin 𝑦 = 𝐶1 cos 𝑥
b) 𝑥(𝑡 2 − 1)𝑑𝑡 + (𝑡 3 − 3𝑡)𝑑𝑥 = 0
Solution:
1
**Multiplying both sides by 𝑥(𝑡 3 −3𝑡) results to,
𝑡2 − 1
𝑑𝑥
∫( 3
) 𝑑𝑡 + ∫
=𝐶
𝑡 − 3𝑡
𝑥
** Reduction
𝑡2 − 1
𝐴 𝐵𝑡 + 𝐶
= + 2
2
𝑡(𝑡 − 3)
𝑡
𝑡 −𝐶
2
2
𝑡 − 1 = 𝐴(𝑡 − 3) + (𝐵𝑡 + 𝐶)𝑡
**By equating coefficients:
For 𝑡 2 :
1 = A + B → (1)
1
For 𝑡 :
c= 0
1
0
For 𝑡 :
-1 = -3A or 𝐴 = 3
Using (1)
1
+𝐵
3
2
𝐵=
3
1=
Hence,
1
2
𝑡+0
3
∫ ( + 32
) 𝑑𝑡 = ln 𝑥 + 𝐶
𝑡
𝑡 −3
1
2 1
ln 𝑡 + ∙ ln(𝑡 2 − 3) + ln 𝑥 = 𝐶
3
3 2
ln 𝑡 + ln(𝑡 2 − 3) + 3 ln 𝑥 = 3𝐶
Or take the inverse natural logarithm of both sides:
3
2
𝑒 ln 𝑥 𝑡 (𝑡 −3) = 𝑒 3𝐶
( 𝑥 3 𝑡 (𝑡 2 − 3) = 𝐶1
,𝑤ℎ𝑒𝑟𝑒 𝐶1 = 𝑒 3𝐶
c) 2𝑥
𝑑𝑦
𝑑𝑥
= 1 + 𝑦 2 ; 𝑤ℎ𝑒𝑛 𝑥 = 2, 𝑦 = 3
2𝑥𝑦 𝑑𝑦 − (1 + 𝑦 2 )𝑑𝑥 = 0
1
Multiplying both sides by 𝑥(1+ 𝑦 2 ) results to:
2𝑦
𝑑𝑥
∫(
) 𝑑𝑦 − ∫
=𝐶
2
1+ 𝑦
𝑥
ln(1 + 𝑦 2 ) − ln 𝑥 = 𝐶
(1 + 𝑦 2 )
ln
=𝐶
𝑥
Take the inverse natural logarithm of both sides:
(1+𝑦 2 ) 𝑐
ln
=𝑒
𝑥
𝑒
(1 + 𝑦 2 )
= 𝐶1 ;
𝑥
𝐶1 = 𝑒 𝑐
For 𝐶1 :
(1 + 32 )
= 𝐶1
2
𝐶1 = 5
Hence,
(1 + 𝑦 2 )
= 5
𝑥
1 + 𝑦 2 = 5𝑥
𝑦 2 = 5𝑥 − 1
𝑦 = √5𝑥 − 1
Quiz #1: Solve the following differential equations.
1. (𝟏 − 𝒙)𝒚′ = 𝒚𝟐
2. 𝒙 𝒄𝒐𝒔𝟐 𝒚 𝒅𝒙 + 𝐭𝐚 𝐧 𝒚 𝒅𝒚 = 𝟎
3. 𝟐𝒚 𝒅𝒙 = 𝟑𝒙𝒅𝒚; 𝒘𝒉𝒆𝒏 𝒙 = 𝟐, 𝒚 = 𝟏
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