Q67.The solution curve of the differential equation, (1 + e−x)(1 + y2) dxdy = y2 which passes through the point (0, 1), is + 2 ) + 2) (1) y2 + 1 = y(loge( 1+e−x2 ) 2) (2) y2 + 1 = y(loge( 1+ex (3) y2 = 1 + y loge( 1+ex2 ) (4) y2 = 1 + y loge( 1+e−x2 )
What This Question Tests
This question tests the ability to solve a first-order differential equation by separating variables and then applying the given initial condition to find the particular solution.
Concepts Tested
Formulas Used
∫ (1/y^2) dy
∫ (1/(1+e^(-x))) dx = ∫ (e^x/(e^x+1)) dx
📚 NCERT Sections This Tests
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.
12.1 — (A) No Different From
Physics Class 12 · Chapter 12
12.1 (a) No different from (b) Thomson’s model; Rutherford’s model (c) Rutherford’s model (d) Thomson’s model; Rutherford’s model (e) Both the models
14.2 — Which Of The Statements Given In Exercise 14.1 Is True For P-Type
Physics Class 12 · Chapter 14
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
📋 Question Details
- Chapter
- Differential Equations
- Topic
- Solution of variable separable differential equations
- Year
- 2020
- Shift
- 03 Sep Shift 1
- Q Number
- Q67
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 9: Differential Equations
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