Differential Equations of First Order & Higher Degree: Lecture 18
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.
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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 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
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 + + + + = 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
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.
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
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
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
