JEE Main 2023 · 12 April · morning · Single correct · +4 / −1
Let y=y(x), y > 0, be a solution curve of the differential equation (1+x²) d y=y(x-y) d x. If y(0)=1 and y(2 √2)=β, then
- (a) e^(β⁻¹)=e⁻²(3+2 √2)
- (b) e^(3 β⁻¹)=e(5+√2)
- (c) e^(3 β⁻¹)=e(3+2 √2)
- (d) e^(β⁻¹)=e⁻²(5+√2)
Answer and solution
Answer (c) e^(3 β⁻¹)=e(3+2 √2)
Solution
Maths sir · 5 steps
The y² term makes this a Bernoulli equation, so put z = 1/y and it turns linear.
- Make it linear. Write it as (1+x²)dy/dx = xy - y² and divide by y². Put z = 1/y, so dz/dx = -1/y²dy/dx. This gives (1+x²)dz/dx + xz = 1.
dz/dx + x/(1+x²) z = 1/(1+x²)
- Integrating factor. The integrating factor is e^(∫ x/(1+x²)dx) = e^(1/2 ln(1+x²)) = √(1+x²). Multiplying through gives d/dx(z√(1+x²)) = 1/√(1+x²), so z√(1+x²) = ln(x+√(1+x²)) + C.
- Use y(0) = 1. At x = 0, z = 1, so 1 = ln 1 + C and C = 1.
- Put x = 2√2. Here √(1+8) = 3 and z = 1/β. So 3/β = ln(3+2√2) + 1.
- So. Take the exponential of both sides: e^(3β⁻¹) = e(3+2√2), which is option (c).
The idea to keep. If a first-order equation has y and y² on the right, divide by y² and substitute z = 1/y. You get a linear equation in z, which an integrating factor solves.
Answer from the NTA final answer key. Solution by PrepWiser.