The solution curve of the differential equation, (1 + e⁻ˣ)(1 + y²) dy/dx = y², which passes through the point (0, 1), is:

(D) y² = 1 + y log e ((1 + eˣ)/2)

(A) y² + 1 = y (log e ((1 + e⁻ˣ)/2) + 2)

(B) y² + 1 = y (log e ((1 + eˣ)/2) + 2)

(C) y² = 1 + y log e ((1 + e⁻ˣ)/2)

Differential Equations: JEE Main 2020 Mathematics question

JEE Main 2020 · 3 Sep, Morning shift · Q3599

JEE Main 20203 Sep, Morning shiftMathematicsMediumDifferential Equations163× since 2002

The solution curve of the differential equation, (1 + e^-x)(1 + y²)

{{dy} \over {dx}}

= y², which passes through the point (0, 1), is :

JEE Main 2020 · 3 Sep, Morning shift · Q3599

JEE Main 20203 Sep, Morning shiftMathematicsMediumDifferential Equations163× since 2002

The solution curve of the differential equation, (1 + e^-x)(1 + y²)

{{dy} \over {dx}}

= y², which passes through the point (0, 1), is :