Q85.The solution of the differential equation ydx −(x + 2y2)dy = 0 is x = f(y). If f(−1) = 1, then f(1) is equal to (1) 2 (2) 3 (3) 4 (4) 1 −−−−−
What This Question Tests
This problem involves solving a first-order linear differential equation. It requires rearranging the equation into the standard form, finding the integrating factor, and then solving for x as a function of y, finally using the initial condition to find the constant C.
Concepts Tested
Formulas Used
dy/dx + P(x)y = Q(x)
Integrating Factor (IF) = e^(∫P(x)dx)
Solution: y * IF = ∫(Q(x) * IF)dx + C
📚 NCERT Sections This Tests
3.9 — A Reaction Is First Order In A And Second Order In B.
Chemistry Class 11 · Chapter 3
3.9 A reaction is first order in A and second order in B. (i) Write the differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times? (iii) How is the rate affected when the concentrations of both A and B are doubled? 85 Chemical Kinetics Reprint 2025-26
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
12.5 — A Hydrogen Atom Initially In The Ground Level Absorbs A Photon,
Physics Class 12 · Chapter 12
12.5 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
📋 Question Details
- Chapter
- Differential Equations
- Topic
- Linear differential equations
- Year
- 2015
- Shift
- 11 Apr Online
- Q Number
- Q85
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 9: Differential Equations
More from this Chapter
Q98.The solution of the differential equation dx dy = x+yx satisfying the condition y(1) = 1 is (1) y = ln x + x (2) y = x ln x + x2 (3) y = xe(x−1) (4) y = x ln x + x
Q99.The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is (1) (x −2)y′2 = 25 −(y −2)2 (2) (y −2)y′2 = 25 −(y −2)2 (3) (y −2)2y′2 = 25 −(y −2)2 (4) (x −2)2y′2 = 25 −(y −2)2 Q100.The non-zero verctors →a,→b and →c are related by →a = 8→b and →c = −7→b. Then the angle between →a and→cis (1) 0 (2) π/4 (3) π/2 (4) π Q101.The vector →a = α^i + 2^j + β^k lies in the plane of the vectors →b = ^i + ^j and →c = ^j + ^k and bisects the angle between →b and →c. Then which one of the following gives possible values of α and β ? (1) α = 2, β = 2 (2) α = 1, β = 2 (3) α = 2, β = 1 (4) α = 1, β = 1 Q102.The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz− plane at the point (0, 172 , −132 ). Then JEE Main 2008 JEE Main Previous Year Paper (1) a = 2, b = 8 (2) a = 4, b = 6 (3) a = 6, b = 4 (4) a = 8, b = 2 Q103.If the straight lines x−1 k = y−22 = z−33 and x−23 = y−3k = z−12 intersect at a point, then the integer k is equal to (1) −5 (2) 5 (3) 2 (4) −2 Q104.It is given that the events A and B are such that P(A) = 41 , P ( BA ) = 12 and P ( BA ) = 32 . Then P(B) is (1) 1 (2) 1 6 3 (3) 2 (4) 1 3 2 Q105.A die is thrown. Let A be the event that the number obtained is greater than 3 . Let B be the event that the number obtained is less than 5 . Then P(A ∪B) is (1) 3 (2) 0 5 (3) 1 (4) 2 5 JEE Main 2008 JEE Main Previous Year Paper
Q86.If →u, →v, ¯w are non-coplanar vectors and p, q are real numbers, then the equality [ 3→u p→v p→w ] −[ p→v →w q→u ] −[ 2→w q→v q→u ] = 0 holds for (1) exactly one value of (p, q) (2) exactly two values of (p, q) (3) more than two but not all values of (p, q) (4) all values of (p, q)
Q84.Solution of the differential equation cos xdy = y(sin x −y)dx, 0 < x < π2 is (1) y sec x = tan x + c (2) y tan x = sec x + c (3) tan x = (sec x + c)y (4) sec x = (tan x + c)y