If the solution curve y=y(x) of the differential equation (1+y²)(1+log ₑ x) d x+x d y=0, x > 0 passes through the point (1,1) and y(e)=(α-tan (3/2))/(β+tan (3/2)), then α+2 β is _________.

Solution

Maths sir · 5 steps

Divide by x(1+y²) and the variables split cleanly; then just match the tan subtraction formula.

1. Separate the variables. Divide the equation by x(1+y²). This gives dy/(1+y²) = -(1+logₑ x)/x dx, with y on one side and x on the other.
2. Integrate both sides. The left side integrates to tan⁻¹ y. On the right, ∫ 1/x dx = logₑ x and ∫ (logₑ x)/x dx = (logₑ x)²/2.

   tan⁻¹ y = -(logₑ x + (logₑ x)²/2) + C

3. Use the point. At (1,1), logₑ 1 = 0, so C = tan⁻¹ 1 = π/4. At x = e, logₑ e = 1, so tan⁻¹ y = π/4 - (1 + 1/2) = π/4 - 3/2.
4. Expand the tangent. y(e) = tan(π/4 - 3/2) = (1 - tan 3/2)/(1 + tan 3/2). Comparing with the given form, α = 1 and β = 1.
5. So. α + 2β = 1 + 2 = 3.

The idea to keep. When 1+y² appears, expect tan⁻¹ y after integrating. An answer shaped like (1-tan A)/(1+tan A) is just tan(π/4-A) in disguise.

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