Tiêu đề 22. DIFFERENTIAL EQUATIONS.pdf ✅

22. DIFFERENTIAL EQUATIONS-MCQS TYPE (1.) The differential equation which represents the family of curves 2 1 c x y c e = , where 1 c and 2 c are arbitrary constants, is [AIEEE-2009] (a.) y y y  =  (b.) yy y  =  (c.) ( ) 2 yy y  =  (d.) 2 y y  = (2.) Solution of the differential equation cos sin ,0 ( ) 2 xdy y x y dx x  = −   is [AIEEE-2010] (a.) sec tan x x c y = + ( ) (b.) y x x c sec tan = + (c.) y x x c tan sec = + (d.) tan sec x x c y = + ( ) (3.) Consider the differential equation 2 1 y dx x dy 0 y   + − =     . If y (1 1 ) = , then x is given by [AIEEE-2011] (a.) 1 1 1 y e y e + − (b.) 1 1 1 y e y e − + (c.) 1 2 4 y e y e − − (d.) 1 1 3 y e y e − + (4.) The population p t( ) at time t of a certain mouse species satisfies the differential equation ( ) 0.5 450 ( ) dp t p t dt = − . If p (0 850 ) = , then the time at which the population becomes zero is [AIEEE-2012] (a.) ln9 (b.) 1 ln18 2 (c.) In 18 (d.) 2ln18 (5.) Let y x( ) be the solution of the differential equation ( log 2 log , 1 , 0 1 ) ( ) ( ) dy x x y x x x y dx + =  = Then y e( ) is equal to [JEE (Main)-2015] (a.) e (b.) 0 (c.) 2 (d.) 2e (6.) If a curve y f x = ( ) passes through the point (1, 1− ) and satisfies the differential equation, y xy dx xdy (1+ = ) , then 1 2 f     −   is equal to [JEE (Main)-2016] (a.) 4 5 − (b.) 2 5 (c.) 4 5 (d.) 2 5 − (7.) If (2 sin 1 cos 0 ) ( ) dy x y x dx + + + = and y (0 1 ) = , then 2 y        is equal to [JEE (Main)- 2017]
(a.) 2 3 − (b.) 1 3 − (c.) 4 3 (d.) 1 3 (8.) let y y x = ( ) be the solution of the differential equation sin cos 4 , 0, ( ) dy x y x x x dx + =   . If 0 2 y    = =     , then 6 y        is equal to [JEE (Main)-2018] (a.) 4 2 9 3  (b.) 8 2 9 3  − (c.) 8 2 9 −  (d.) 4 2 9 −  (9.) If y y x = ( ) is the solution of the differential equation, 2 2 dy x y x dx + = satisfying y (1 1 ) = , then 1 2 y       is equal to [JEE (Main)-2019] (a.) 13 16 (b.) 7 64 (c.) 1 4 (d.) 49 16 (10.) Let f R : 0,1   → be such that f xy f x f y ( ) =  ( ) ( ) , for all x y, 0,1   , and f (0 0 )  . If y y x = ( ) satisfies the differential equation, ( ) dy f x dx = with y (0 1 ) = , then 1 3 4 4 y y         +     is equal to [JEE (Main)-2019] (a.) 4 (b.) 3 (c.) 2 (d.) 5 (11.) If 2 2 3 1 , , cos cos 3 3 dy y x dx x x   −  + =     and 4 4 3 y      =   , then 4 y      −   equals [JEE (Main)-2019] (a.) 1 6 3 + e (b.) 1 3 3 + e (c.) 1 3 (d.) 4 3 − (12.) The curve amongst the family of curves represented by the differential equation, ( ) 2 2 x y dx xydy − + = 2 0 which passes through (1,1) is [JEE (Main)-2019] (a.) A hyperbola with transverse axis along the x -axis. (b.) A circle with centre on the y -axis. (c.) An ellipse with major axis along the y -axis. (d.) A circle with centre on the x -axis. (13.) If y x( ) is the solution of the differential equation 2 1 2 , 0 dy x x y e x dx x   + − + =      , where ( ) 1 2 1 2 y e − = , then [JEE (Main)-2019]
(a.) y x( ) is decreasing in 1 ,1 2       (b.) ( ) log 2 log 2 4 e e y = (c.) y (log 2 log 4 e e ) = (d.) y x( ) is decreasing in (0,1) (14.) The solution of the differential equation, 2 ( ) dy x y dx = − , when y (1 1 ) = , is [JEE (Main)-2019] (a.) ( ) 2 log 2 1 2 e y y x − = − − (b.) 0 1 log 2 1 x y x y x y + − − = + − − + (c.) 0 ( ) 1 log 2 1 1 x y x x y − + − = − + − (d.) 6 2 log 2 x x y y − = − − (15.) Let y y x = ( ) be the solution of the differential equation, log ,( 1) e dy x y x x x dx + =  . If 2 2 y ( ) = log 4 1 e − , then y e( ) is equal to [JEE (Main)-2019] (a.) 2 4 e (b.) 2 e − (c.) 2 2 e − (d.) 4 e (16.) Let y y x = ( ) be the solution of the differential equation, ( ) ( ) 2 2 2 1 2 1 1 dy x x x y dx + + + = such that y (0 0 ) = . If (1) 32 a y  = , then the value of ' a ' is [JEE (Main)-2019] (a.) 1 2 (b.) 1 4 (c.) 1 (d.) 1 16 (17.) Given that the slope of the tangent to a curve y y x = ( ) at any point ( x y, ) is 2 2y x . If the curve passes through the centre of the circle 2 2 x y x y + − − = 2 2 0 , then its equation is [JEE (Main)-2019] (a.) x y x log 2 1 c = − − ( ) (b.) log 1 e x y x = − (c.) x y x log 2 1 e = − ( ) (d.) ( ) 2 log 2 1 e x y x = − − (18.) The solution of the differential equation ( ) 2 2 0 dy x y x x dx + =  with y (1 1 ) = , is [JEE (Main)-2019] (a.) 3 2 4 1 5 5 y x x = + (b.) 3 2 1 5 5 x y x = + (c.) 2 2 3 4 4 x y x = + (d.) 2 2 3 1 4 4 y x x = +
(19.) If y y x = ( ) is the solution of the differential equation ( ) 2 tan sec , , 2 2 dy x y x x dx     = −  −    , such that y (0 0 ) = , then 4 y      −   is equal to [JEE (Main)-2019] (a.) 1 2 e − (b.) 1 2 − e (c.) e−2 (d.) 1 2 e + (20.) Let y y x = ( ) be the solution of the differential equation, 2 tan 2 tan , , 2 2 dy y x x x x x dx     + = +  −    , such that y (0 1 ) = . Then [JEE (Main)-2019] (a.) 2 2 4 4 2 y y            + − = +     (b.) 2 4 4 y y           − − =     (c.) 2 4 4 y y            + − = −      (d.) 2 4 4 y y            − − = −       (21.) If y e xy e + = , the ordered pair 2 2 , dy d y dx dx       at x = 0 is equal to [JEE (Main)-2019] (a.) 2 1 1 , e e     −   (b.) 2 1 1 , e e     − −   (c.) 2 1 1 , e e     −   (d.) 2 1 1 , e e       (22.) Consider the differential equation, 2 1 y dx x dy 0 y   + − =     . If value of y is 1 when x =1 , then the value of x for which y = 2 , is [JEE (Main)-2019] (a.) 5 1 2 e + (b.) 3 2 − e (c.) 3 1 2 e − (d.) 1 1 2 e + (23.) The general solution of the differential equation ( ) ( ) 2 3 y x dx xydy x − − =  0 0 is (where c is a constant of integration) [JEE (Main)-2019] (a.) 2 3 2 y x cx + + = 2 0 (b.) 2 2 3 y x cx − + = 2 0 (c.) 2 3 2 y x cx − + = 2 0 (d.) 2 2 3 y x cx + + = 2 0 (24.) Let ,( , 0) k k k x y a a k + =  and 1 3 0 dy y dx x   + =     , then k is [JEE (Main)-2020] (a.) 3 2 (b.) 1 3 (c.) 4 3 (d.) 2 3 (25.) If y y x = ( ) is the solution of the differential equation, 1 y x dy e e dx     − =   such that y (0 0 ) = , then y (1) is equal to [JEE (Main)-2020]