Differential Equations of First Order & Higher Degree: Lecture 18

 
Differential Equation
 Lecture-18
Differential Equation of the first order
and higher degree
UG (B.Sc., Part-2)
 
Dr. Md. Ataur Rahman
Guest Faculty
Department of Mathematics
M. L. Arya, College, Kasba
PURNEA UNIVERSITY, PURNIA
 
Contents
 
Differential equation of the first order but not
of the first degree
The general form of differential equation of the
first order but not of the first degree
i.e. the general form of differential equation of
the first order and higher degree.
 Solution of the first order and  higher degree
D.E.
 
Differential equation of the first
order but not of the first degree
 
The general form of a differential equation of the
first order and nth degree is
 
Solution of the first order and higher
degree D.E.
 
Methods of solving such types of differential
equations
1.
Equations Solvable for 
p
2.
Equations Solvable for 
y
3.
Equations Solvable for 
x
 
1. Equations Solvable for 
p
 
Let the general form of the differential equation
of the first order and nth degree be
 
Working rule:- 
Factorize (1) into n linear factors
i.e.
Equating each factor of (2) to zero, we get
 
 
Continue
 
After integrating ,We get
 
Then the general solution of (1) is
 
 
Since the equation (1) is the first order differential
equation.
So, the general solution of (1) contains only one
arbitrary constant.
 
 
Problems
 
Solve the following differential equations
 
Solution
 
Solution (1):- Given equation is
 
Continue
 
Integrating it, we get
 
 
 
 
Then the general solution of (1) is
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This lecture covers differential equations of first order but not of the first degree, general forms of such equations, methods for solving them, and examples of differential equations to be solved. The content includes detailed explanations, equations, solutions, and problem-solving techniques.

  • Differential Equations
  • First Order
  • Higher Degree
  • Solutions
  • Mathematics

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  1. Differential Equation Lecture-18 Differential Equation of the first order and higher degree UG (B.Sc., Part-2) Dr. Md. Ataur Rahman Guest Faculty Department of Mathematics M. L. Arya, College, Kasba PURNEA UNIVERSITY, PURNIA

  2. Contents Differential equation of the first order but not of the first degree The general form of differential equation of the first order but not of the first degree i.e. the general form of differential equation of the first order and higher degree. Solution of the first order and higher degree D.E.

  3. Differential equation of the first order but not of the first degree The general form of a differential equation of the first order and nth degree is 1 2 n n n + dy dx dy dx dy dx + + + + = 0 Q Q Q 1 2 n + + + = 1 2 n n n . . . . ie 0 ie p Q p Q p Q 1 2 n = ( , , f x y p ) 0 dy dx = , , , tan . Where p andQ Q Q Q arecons tsor functionsof xand y 1 2 3 n

  4. Solution of the first order and higher degree D.E. Methods of solving such types of differential equations 1. Equations Solvable for p 2. Equations Solvable for y 3. Equations Solvable for x

  5. 1. Equations Solvable for p Let the general form of the differential equation of the first order and nth degree be + + + + = 1 2 n n n 0... (1) p Q p Q p Q 1 2 n Working rule:- Factorize (1) into n linear factors i.e. Equating each factor of (2) to zero, we get = = ( , ) . f x y ( , ) . f x y ( , ) f x y ( , ) f x y 0... (2) p p p p 1 2 3 n = = ( , ) f x y dy dx 0, ( , ) f x y 0, ( , ) f x y 0 p p p 1 2 2 dy dx dy dx = = = ( , ), f x y ( , ), f x y ( , ) x y f 1 2 n

  6. Continue After integrating ,We get ( , , ) x y c = = = 0, ( , , x y c ) 0, ( , , x y c ) 0 1 1 2 2 n n Then the general solution of (1) is ( , , ) x y c c c = = = = ( , , ... = ) = ( , , ) x y c 0 x y c 1 2 n ( ) c c c say 1 2 3 n Since the equation (1) is the first order differential equation. So, the general solution of (1) contains only one arbitrary constant.

  7. Problems Solve the following differential equations + + + + = 2 2 1. 2. 3. 2 7 2 3 0 p p p xp p py x = 2 12 cot 0 = 2 2 x y 2 + dy dx dy dx = = 4. ( ) 0 x y x y 2 x x 5. 6. ( x ) + 0 ) p p p p e + e y x = ( ) ( y

  8. Solution Solution (1):- Given equation is + = 2 2 2 3 0 p xp + x = 2 2 3 + 3 0 = p p p p p xp xp x p x + ( 3 ) x x x ( x or p dy dx 3 ) 0 0 x + + = ( ( dy dx dy 3 )( 3 ) ) p = = 0 ( ) 0 x = = 3 x or x = = 3 xdx or dy xdx

  9. Continue Integrating it, we get 2 2 3 2 x x x = + = + y c or y c 2 2 2 3 x + = = 0 0 y c or y c 2 3 = 2 + = = 2 2 2 0 2 0 y x k or y x k 2 wherek c Then the general solution of (1) is + = 2 2 (2 3 )(2 ) 0 y x k y x k

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