If y = y(x) is the solution of the differential equation ( 1 + e²ˣ )dy/dx + 2( 1 + y² )eˣ = 0 and y (0) = 0, then 6( y'(0) + ( y( log ₑ√3 ) )² ) is equal to

Solution

Maths sir · 5 steps

Separate the variables and both sides become arctan integrals. The constant comes from y(0) = 0.

1. Separate variables. Move all y terms to one side and all x terms to the other: dy/(1+y²) = -2eˣ/(1+e²ˣ) dx. On the right, put t = eˣ, so dt = eˣ dx and the integral becomes ∫ (-2 dt)/(1+t²).

   tan⁻¹ y = -2 tan⁻¹(eˣ) + C

2. Find the constant. At x = 0, y = 0 and e⁰ = 1, so 0 = -2 × π/4 + C. That gives C = π/2, and so tan⁻¹ y = π/2 - 2 tan⁻¹(eˣ).
3. Find y'(0). Put x = 0 and y = 0 straight into the given equation: (1+1) y'(0) + 2(1+0)(1) = 0. So y'(0) = -1.
4. Find y at log√3. At x = logₑ√3, eˣ = √3 and tan⁻¹√3 = π/3. So tan⁻¹ y = π/2 - 2π/3 = -π/6, which gives y = -1/√3 and y² = 1/3.
5. So. 6(-1 + 1/3) = 6 × (-2/3) = -4. The answer is (c).

The idea to keep. To find y' at a point, put the known x and y values straight into the differential equation. You do not need the full solution for that. Keep the full solution for finding y at other points.

Answer from the NTA final answer key. Solution by PrepWiser.