Q76.The solution of the differential equation (x2 + y2)dx −5xy dy = 0, y(1) = 0, is : (1) x2 −2y2 6 = x (2) x2 −4y2 6 = x (3) x2 −4y2 5 = x2 (4) x2 −2y2 5 = x2 →
What This Question Tests
This question involves solving a first-order differential equation using the variable separable method, identifying the resulting curve as a conic section (parabola), and then finding the line on which its vertex lies.
Concepts Tested
Formulas Used
∫ dy / (Ay+B) = ∫ dx / (Cx+D)
y^2 = 4ax (parabola equation)
📚 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.
9.15 — Apply Mirror Equation And The Condition:
Physics Class 12 · Chapter 9
9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
10.2 — What Is The Shape Of The Wavefront In Each Of The Following Cases:
Physics Class 12 · Chapter 10
10.2 What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.
📋 Question Details
- Chapter
- Differential Equations
- Topic
- Solving differential equations, equation of conic sections
- Year
- 2024
- Shift
- 09 Apr Shift 1
- Q Number
- Q76
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 9: Differential Equations
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