JEE Main 2020 · 8 January · evening · Single correct · +4 / −1
The differential equation of the family of curves, x² = 4b(y + b), b ∈ R, is :
- (a) x(y')² = x – 2yy'
- (b) x(y')² = 2yy' – x
- (c) xy" = y'
- (d) x(y')² = x + 2yy'
Answer and solution
Answer (d) x(y')² = x + 2yy'
Solution
Maths sir · 4 steps
One constant b means one differentiation: find b from y', then put it back to eliminate it.
- Differentiate once. The family has one arbitrary constant, so the equation is first order. Write the curve as x² = 4by + 4b² and differentiate with respect to x: 2x = 4by', so b = x/2y'.
- Eliminate b. Put this b back into x² = 4by + 4b². That gives x² = 2xy/y' + x²/(y')².
x² (y')² = 2xy y' + x²
- Tidy up. The line above comes from multiplying through by (y')². Now divide by x, which is non-zero for the curves of interest: x(y')² = 2yy' + x.
- So. The differential equation is x(y')² = x + 2yy', which is option (d).
The idea to keep. Count the arbitrary constants first: that is the order. Differentiate that many times, solve for the constants, and substitute them back into the original equation.
Answer from the NTA final answer key. Solution by PrepWiser.