Solving Various Differential Equations with Step-by-Step Solutions
Explore detailed solutions to differential equations with step-by-step explanations for equations involving trigonometric functions, exponentials, and derivatives. Enhance your understanding of differential equations through comprehensive problem-solving techniques.
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Differential Equation Lecture-27 Problems Dated 01.05.2020 PPT-01 UG (B.Sc., Part-2) Dr. Md. Ataur Rahman Guest Faculty Department of Mathematics M. L. Arya, College, Kasba PURNEA UNIVERSITY, PURNIA
Problems Solve the following Equations 2 2 d y dx d y dx d y dx dy dx dy dx dy dx d y dx d y dx dy dx + + = + = 2 2 3 x x (1). 3 2 (2) 2 y e y x e 2 2 2 2 dy dx + = + = + + 2 2 (3). 4 4 (4). 2 4 y x x x 2 2 2 2 d y dx + = + = 2 (5) 5 6 3 (6) y Sin x a y Cosax 2 2 2 d y dx + = = 3 2 (7) 4 2 (8) D y Dy Sin x y x Cosx 2 2 2 d y dx dy dx d y dx + = + + = 2 2 x ax (9) 4 4 2 (1 0) y x e Cos x a y e 2 2
Solution of (2) Solution (2):- Given Eq. is 2 d y dx dy dx + = 2 3 x e x 2 ....(1) y 2 ( ) + = 2 2 3 x e x 2 1 D D y The auxiliary Eq. is + = + = 2 2 2 1 1 0 0 m m m m 2 ( m ) 2 = 1 0 m = 1,( = ), twice c + x . ( ) C F c x e 1 2
Continue (2) 1 2 1 Now, = = 2 3 x e 2 3 x e x x . P I ( ) 1 + + 2 2 1 D D 1 D 1 3 1 + = = 3 2 3 2 x x e x e x ( ) ( ) 2 2 2 D D 2 2 3 3 x 11 4 D e D D D = + = 1 2 + + 3 2 2 x 3 4 e x x 2 4 2 4 8 3 3 x x 3 4 e e ( ) = + 2 0 + = + 2 2 2 2 4 3 x x x x 4 8 Then the complete Solution of (1) is 3 x e ( ) = + = + + + 2 x . . ( ) 2 4 3 y C F P I c c x e x x 1 2 8
Solution of (3) Solution (3):- Given Eq. is 2 d y dx dy dx + = 2 4 4 ....(1) y x 2 ( ) + = 2 2 4 4 D D y x The auxiliary Eq. is + + = = 2 4 4 4 4 0 0 m m m m 2 ( m ) 2 = 2 0 m = 2,( ( = ), twice c + 2 x . ) C F c x e 1 2
Continue (3) 1 4 1 Now, = = 2 2 . P I x x ( ) + 2 2 4 D D 2 D 2 2 3 1 4 1 4 D D D D = = 1 2 + + + + 2 2 1 3 4 x x 2 2 4 8 + + 2 1 4 3 4 1 4 4 2 3 1 8 x x ( ) = + + = = + + 2 2 2 2 4 3 x x x x Then the complete solution of (1) is 1 8 ( ) = + = + + + + 2 2 x . . ( ) 4 3 y C F P I c c x e x x 1 2
Solution of (5) Solution (5):- Given Eq. is 2 d y dx dy dx + = 5 6 3 ....(1) Sin x y 2 ( ) + = 2 5 6 3 D D y Sin x The auxiliary Eq. is + + = = 2 5 5 6 6 0 0 m m m m = 2 ( m )( ) = 2 3 0 m m 2, 3 c e = + 2 3 x x . C F c e 1 2
Continue (5) Now, 1 5 = . 3 P I Sin x + 2 6 D D 1 5 1 = = 3 3 Sin x Sin x + 2 3 6 3 5 D D D D 1 5 3 = = 3 3 Sin x Sin x 3 5 + 2 25 9 5 D 5 3 3 D D = = 3 3 Sin x Sin x 2 2 25( 3 ) 9 1 15 234 25( 3 ) 9 1 78 ( ) ( ) = = 3 3sin3 5 3 sin3 Cos x x Cos x x Then the complete solution of (1) is 1 78 ( ) = + = + + 2 3 x x . . 5 3 sin3 y C F P I c e c e Cos x x 1 2
Solution of (6) Solution (6):- Given Eq. is 2 d y dx + = 2 ....(1) a y Cosax 2 ( ) + = 2 2 D a y Cosax The auxiliary Eq. is + = 2 2 0 m a ( ) ai ai 2 = 2 0 m ( )( ) + = 0 m m ai = m ai = + . C F ACosax BSinax
Continue (6) + iax iax 1 + 1 1 e e Now, = = = . P I Cosax Cosax ( )( ) ( ) ai + 2 2 2 2 D a D ai D ai 2 D 1 1 = + iax iax e e ( )( ) ( )( ) + + 2 2 D ai D ai D ai D ai 1 1 = + iax iax e e ( )( ) ( )( ) + + 2 2 ai ai D ai D ai ai ai 1 1 = iax iax e e ( 1 4 ) ( ) 1 ai + 4 4 ai D ai ai D ai = aix aix aix aix aix a ix e e e dx e e e dx 4 ai 1 ai x ai 1 ai 1 ai x a 1 ai = = aix aix aix aix e dx e dx e x e x 4 4 4 4 x ai ( ) = = = aix aix 2 e e iSinax Sinax 4 4 2 Then the complete solution of (1) is x a = + = + + . . y C F P I ACosax BSinax Sinax 2
Solution of (10) Solution (10):- Given Eq. is 2 d y dx = 2 ax ....(1) a y e 2 ( ) = 2 2 ax D a y e The auxiliary Eq. is = 2 2 0 m a ( )( ) + = 0 m a m a = m a c e = + ax ax . C F c e 1 2
Continue (10) Now, 1 1 = = ax ax . P I e e ( )( ) D a + D a 2 2 D a 1 1 = = = = ax ax 0 e e as D a for D a ( )( ) ( ) a a + D a 2 a D a ax 1 1 xe = = = ax ax ax ax e e e dx e dx 2 2 2 a a a Then the complete solution of (1) is 1 = + = + + ax ax ax . . y C F P I ce c e xe 1 2 2 a
