JEE Main 2020 · 3 September · morning · Single correct · +4 / −1
The solution curve of the differential equation,
(1 + e⁻ˣ)(1 + y²)dy/dx = y²,
which passes through the point (0, 1), is :
- (a) y² + 1 = y( log ₑ( (1 + e⁻ˣ)/2 ) + 2 )
- (b) y² + 1 = y( log ₑ( (1 + eˣ)/2 ) + 2 )
- (c) y² = 1 + ylog ₑ( (1 + e⁻ˣ)/2 )
- (d) y² = 1 + ylog ₑ( (1 + eˣ)/2 )
Answer and solution
Answer (d) y² = 1 + ylog ₑ( (1 + eˣ)/2 )
Solution
Maths sir · 4 steps
Split the y-bits and x-bits, and multiply top and bottom by eˣ before you integrate the right side.
- Separate variables. Move every y to the left and every x to the right: (1+y²)/y² dy = dx/(1+e⁻ˣ). Multiply the right side's top and bottom by eˣ to get eˣ/(eˣ+1) dx.
- Integrate both sides. On the left, (1+y²)/y² = 1/y² + 1, which integrates to -1/y + y. On the right, the top is the derivative of the bottom, so it integrates to ln(1+eˣ).
y - 1/y = ln(1+eˣ) + C
- Use the point. At (0, 1): 1 - 1 = ln 2 + C, so C = -ln 2. This gives y - 1/y = ln((1+eˣ)/2).
- So. Multiplying through by y gives y² - 1 = y ln((1+eˣ)/2), that is, y² = 1 + y ln((1+eˣ)/2). The answer is (d).
The idea to keep. When you see 1/(1+e⁻ˣ), multiply top and bottom by eˣ. The integrand becomes f'/f and integrates straight to a log.
Answer from the NTA final answer key. Solution by PrepWiser.