Tiêu đề MP DPP Sheet 03.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 2025 (Online Batch) SECTION: MATHEMATICAL PHYSICS Daily Practice Problem (DPP) Sheet 3: DIFFERENTIAL EQUATIONS (Linear Homogeneous ODE with constant co-efficients) Q.1. Solve the differential equations: (i) 2 2 5 6 0 d y dy y dx dx    given that 0 (0) 1, 0 x dy y dx     [Ans. 3 2 2 3 x x y e e   ] (ii) 2 2 6 13 0 d y dy y dx dx    given that 0 (0) 2, 1 x dy y dx    [Ans. 3 7 2cos2 sin 2 2 x y e x x         ] (iii) 2 2 8 16 0 d y dy y dx dx    given that 0 (0) 1, 2 x dy y dx    [Ans.   4 1 2 x y x e   ] (iv) 2 2 2 17 0 d y dy y dx dx    given that 4 (0) 1, 0 dy y dx    [Ans. 1 cos 4 sin 4 4 x y e x x         ] (v) 3 2 3 2 4 4 0 d y d y dy y dx dx dx     given that       ' " y y y 0 0, 0 0, 0 5     [Ans. 1 cos 2 sin 2 2 x    e x x ] (vi) 3 2 3 2 6 11 6 0 d y d y dy y dx dx dx     given that       ' '' y y y 0 0, 0 1, 0 1     [Ans. 2 3 2 3 x x x e e e      ] PART - A: MULTIPLE CHOICE QUESTIONS (MCQ) Q.2. If 2t y e  is a solution to 2 2 5 0 d y dy ky dt dt    , what is the value of k? (a) 2 (b) 4 (c) 6 (d) -6 Q.3. The general solution to 2 2 6 9 0 d z dz z dt dt    is z. Which of the following options are correct? (where A and B are arbitary constants) (a) As t z A    , for any value of B (b) The behaviour of z as t   depends on A, B (c) As t z    , 0 for any value of A and B (d) As t z     , for any value of A and B
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 Q.4. The solution of the differential equation '' y y   4 0 subjected to the condition y 0 1,     ' y 0 2  is (a) sin 2 cos 2 x x  (b) sin 2 cos 2 x x  (c) 1 sin 2 cos 2 2 x x  (d) 1 sin 2 cos 2 2 x x  Q.5. The solution of the differential equation 2 2 2 0 d x dx x dt dt    , which satisfies x 0 3   and does not blow up as t  , will be (a) 3 t x e   (b) 2 4 t t e e x    (c) 2 3 t x e  (d) 2 2 t t x e e     Q.6. The general solution of the differential equation 5 4 3 2 5 4 3 2 3 3     0 d y d y d y d y dx dx dx dx is (a)   2 1 2 3 4 5 x c c x c c x c x e     (b) 2 2 3 1 2 3 4 5     x x x c x c x c e c e c e (c)  1 2 3 4 5      x c c c c c e (d) None of these Q.7. The general solution of the differential equation 3 2 3 2 9 27 27 0 d y d y dy y dx dx dx     is (a)     2 3x y x A Bx Cx e    (b)     2 3x y x A Bx Cx e    (c)   3 3 3 x x x y x Ae Be Ce    (d)   3 3 3 x x x y x Ae Be Ce       Q.8. The general solution of the differential equation 4 4 4 d y m y dx  is (a)   1 2 3 4 y x C mx C mx C mx C mx     cos sin cosh sinh (b)   1 2 3 4 cos sin   mx y x C mx C mx C C x e     (c)     2 3 1 2 3 4 mx y x C C x C x C x e     (d)    1 2 3 4  cosh sinh mx y x C C x e C mx C mx     Q.9. The solution of the ordinary differential equation 2 2 0 d y dy p qy dx dx    is 3 1 2 x x y C e C e     . The value of p and q are respectively (a) 3, 3 (b) 3, 4 (c) 4, 3 (d) 4, 4 Q.10. If 6 and -9 are the roots of the auxiliary equation, then the differential equation can be represented as (a) 2 2 4 12 0 d y dy y dx dx    (b) 2 2 4 12 0 d y dy y dx dx    (c) 2 2 3 54 0 d y dy y dx dx    (d) 2 2 3 54 0 d y dy y dx dx   
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. A function n(x) satisfies the differential equation     2 2 2 0 d n x n x dx L   , where L is a constant and subjected to the conditions n K 0  and n  0 . The solution of the equation will be (a)   x L/ n x Ke  (b)   x L/ n x Ke  (c)   x L / n x Ke  (d)   x L / n x Ke  Q.12. The solution to the differential equation 2 2 d u du k dx dx  , where k is constant, subjected to the boundary conditions     0 u u L u 0 0,   , is (a) 0 x u u L  (b) 0 1 1 kx kL e u u e          (c) 0 1 1 kx kL e u u e            (d) 0 1 1 kx kL e u u e          Q.13. Consider the differential equation: 2 2 0 d y y dt   with initial conditions y 0 1,   0 1 t dy dt   for 0 t  . The solution attains a maximum for t equal to (a) 4 t   (b) 2 t   (c) t  0 (d) t   Q.14. The differential equation 2 2 0 d y y dx   satisfying the conditions y y 0 1, 0      , has (a) a unique solution (b) infinite no.of solutions (c) no solution (d) none of these Q.15. A solution of the ordinary differential equation 2 2 5 6 0 d y dy y dx dx    is such that     3 3 1 0 2 and 1 e y y e    . The value of   0 dy dx is (a) 1 (b) e (c) e2 (d) -3 Q.16. If the characteristic equation of the differential equation 2 2 2 0 d y dy y dx dx     has two equal roots, then the values of  are (a) +1, -1 (b) 0, -1 (c) 0, 1 (d) +1/2, -1/2 Q.17. Consider the following differential equation:       2 2 8 6 0 with 0 1and Lim 0 x d y i y y y x dx       The value of y 2  is (a) 6 e  (b) 6 e   (c) 6 e   (d) 6 e   
4 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 PART - B: Numerical Answer Type (NAT) Questions Q.18. Consider the differential equation: 2 2 2 0 d x dx x dt dt    with initial conditions x x 0 0 and 0 1      . The solution x t  attains its maximum value when ‘t’ is ________________________________________ [Your answer should be upto TWO DECIMAL PLACES] Q.19. Consider the differential equation 2 2 2 0 d x dx x dt dt    . At time t = 0, it is given that x = 1 and 0 dx dt  . At t = 1, the value of x is ____________________________ [Your answer should be an INTEGER] Q.20. Consider the following differential equation:     2 2 6 8 0 with 0 0 and ' 0 1 d y dy y y y dx dx      The minimum value of the solution y x  is _____________________________________________ [Your answer should be upto TWO DECIMAL PLACES] Answer Key PART - A: MULTIPLE CHOICE QUESTIONS (MCQ) 2. (c) 3. (c) 4. (a) 5. (c) 6. (a) 7. (a) 8. (a) 9. (c) 10. (d) 11. (b) 12. (b) 13. (a) 14. (c) 15. (d) 16. (a) 17. (c) PART - C: Numerical Answer Type (NAT) Questions 18. (1) 19. (0.73 to 0.75) 20. (-0.12 to -0.13)