Tiêu đề IIT-JAM Differential Equation DPP Sheet 01.pdf ✅

1 North Delhi : 56-58, First Floor, Mall Road, G.T.B. Nagar (Near Metro Gate No. 3), Delhi-09, Ph: 011-41420035 South Delhi : 28-A/11, Jia Sarai, Near-IIT Metro Station, New Delhi-16, Ph : 011-26851008, 26861009 IIT JAM PHYSICS 2026 (Online Batch) SECTION: MATHEMATICAL PHYSICS Daily Practice Problem (DPP) Sheet D1: DIFFERENTIAL EQUATIONS (1st order Seperable ODE, Reducible to Seperable ODE) Q.1. Solve the following differential equations: (i) 2 2 x y dx y x dy 1 1 0     given that y(0) 3  [Ans. 2 2 1 1 3     x y ] (ii) ( 1)cos sin 0 y y e xdx e xdy    given that 0 3 y         [Ans. sin 1 3   y x e   ] (iii) ln 2 dy x x y dx  given that    2 y 2 ln 2  [Ans.  2 y x  ln ] (iv) 1 1 x dx e dt         given that x 0 1   [Ans. 1 1  x t e e e    ] (v)     2 2 2 given that 0 1 6 x x e dy y dx y    [Ans. 2 3 2 4 2 3 x y e x    ] Q.2. Solve the following differential equations: (i) 2 (4 1) dy x y dx    given that y 0 1   [Ans. 1 4 1 tan 2 2 4 x y x             ] (ii)     4 3 sec given that 1 6 dy x x y xy y dx      [Ans.   2 1 sin 2 xy x  ] (iii) 1 y x dy e e dx         given that y 1 1   [Ans.   1 2 2 2 x y x e e e    ] Q.3. Solve the following differential equations: (i) sin sin 0 y y y x dx x dy x x          given that y 4   [Ans. 1 cos ln 4 ln 2 y x x          ] (ii) (1 ) 1 0 x y x y x e dx e dy y           given that y 0 1   [Ans. x y/ x 1 e y y   ] (iii) 2 2 dy y x y dx x    given that y 1 0   [Ans. 2 2 2 y x y x    ]
2 North Delhi : 56-58, First Floor, Mall Road, G.T.B. Nagar (Near Metro Gate No. 3), Delhi-09, Ph: 011-41420035 South Delhi : 28-A/11, Jia Sarai, Near-IIT Metro Station, New Delhi-16, Ph : 011-26851008, 26861009 (iv) 2 2 2 dy x xy x y dx    given that y 1 0   [Ans. ln 1 x x x y    ] (v)     3 3 2 x y dx xy dy y     3 0 given that 1 0 [Ans. 3 3 x y x   2 ] PART - A: MULTIPLE CHOICE QUESTIONS (MCQ) Q.4. The degree and order of differential equation 3 3/2 2 2 2 4               dy d y dx dx are respectively (a) order 2, degree 3 (b) order 1, degree 3 (c) order 3, degree 2 (d) order 3, degree1 Q.5. What is the degree of the differential equation 3 sin dy dy y x dx dx         ? (a) 1 (b) 2 (c) 8 (d) not defined Q.6. The order and degree of the differential equation 5/3 2 2 2 1 dy d y a dx dx                        are respectively (a) 2, 1 (b) 2, 3 (c) 2, 5 (d) 1, 2 Q.7. The differential equation 3 2 3 2 0 2 d f f d f dx dx   is a (a) second order non-linear ODE (b) third order linear ODE (c) third order non-linear ODE (d) mixed order non-linear ODE Q.8. tan–1 x + tan–1y = c is the general solution of the differential equation (a) 2 2 1 1    dy y dx x (b) 2 2 1 1    dy x dx y (c)     2 2 1 1 0     x dy y dx (d)     2 2 1 1 0     x dx y dy Q.9. The differential equation for which solution is given as x x xy Ae Be   , will be (a) 2 2 2 0 d y dy x x xy dx dx    (b) 2 2 2 0 d y dy x x xy dx dx    (c) 2 2 2 0 d y dy x xy dx dx    (d) 2 2 2 0 d y dy x xy dx dx    Q.10. The size of a popoulation ‘P’ is modeled by the differential equation 1.2 1 4200 dP P P dt         . For what values of P, population is increasing? (a) P > 0 (b) 0 < P < 4200 (c) P < 2100 (d) Not enough information provided.
3 North Delhi : 56-58, First Floor, Mall Road, G.T.B. Nagar (Near Metro Gate No. 3), Delhi-09, Ph: 011-41420035 South Delhi : 28-A/11, Jia Sarai, Near-IIT Metro Station, New Delhi-16, Ph : 011-26851008, 26861009 Q.11. The differential equation of all circles passing through the origin and having their centres on the y-axis (a)   2 2 x y dy dx xy   / 2 (b)   2 2 x y dy dx xy   / 2 (c)   2 2 dy dx xy x y / 2   (d)   2 2 dy dx xy x y / 2   Q.12. The particular solution of differential equation   2 sin 3 dy x x dx  given y 0 0   , is (a)   2 cos 3 6 x y   (b)   2 cos 3 1 6 x y   (c)   2 cos 3 1 6 x y    (d)   2 1 cos 3 6 x y   Q.13. Biotransformation of an organic compound having concentration (x) can be modelled using the ordinary differ- ential equation 2 0 dx kx dt   , where k is the reaction rate constant. If x = a at t = 0, the solution will be (a) kt x ae  (b) 1 1 kt x a   (c) 1  kt x a e   (d) x a kt   Q.14. Consider the following differential equation:   3 2 such that 0 1 1 dy xy y dx x     The range of x for which the solution will be real, will be (a) 3 3 2 2    x (b) 5 5 2 2    x (c)    2 2 x (d) 0 2  x Q.15. The solution of the differential equation  2 2 y x dx xydy    2 0 will be represented by (a) Family of circles having center at the origin. (b) Family of circles having center on x-axis and touching the origin (c) Family of circles having center on y-axis and touching the origin (d) Family of circles having center on y-axis but not touching the origin Q.16. The geometrical interpretation of the graph of the differential equation:   dy x dx y is a family of (a) Hyperbola (b) Ellipse (c) Parabola (d) Circle Q.17. A drop of liquid evaporates at a rate proportional to it’s area of surface. If the radius of the drop is initially 4 mm and 5 minutes later, the radius is reduced to 2 mm, then the radius of the drop after 10 minutes will be (a) 0 mm (b) 0.5 mm (c) 0.8 mm (d) 1 mm Q.18. Which of the following graph best represents the function y f x  ( ) , where f x( ) is the solution of differential equation dy dx xy / 2   with y(0) 2  is (a) 0 x y 2 (b) 0 x y 1 (c) –2 x y (d) –1 x y