If the solution curve of the differential equation (y-2 log ₑ x) d x+(x log ₑ x²) d y=0, x > 1 passes through the points (e, 4/3) and (e⁴, α), then α is equal to ____________.

Solution

Maths sir · 4 steps

Rewrite logₑ x² as 2 logₑ x first, and this becomes a clean linear equation in y.

1. Make it linear. Since ln x² = 2 ln x, the equation is (y - 2 ln x) dx + 2x ln x dy = 0. Divide by 2x ln x dx (allowed since x > 1) to get dy/dx + y/(2x ln x) = 1/x.
2. Integrating factor. With u = ln x, ∫ dx/(2x ln x) = 1/2 ln(ln x). So the integrating factor is e^(1/2 ln(ln x)) = √(ln x).

   y√(ln x) = ∫ √(ln x)/x dx = 2/3(ln x)^(3/2) + C

3. Find C. At (e, 4/3) we have ln x = 1, so 4/3 = 2/3 + C, which gives C = 2/3.
4. So. At x = e⁴, ln x = 4 and √(ln x) = 2. Then 2α = 2/3(8) + 2/3 = 6, so α = 3.

The idea to keep. Whenever x ln x sits under y, expect the integrating factor to be a power of ln x. Substitute u = ln x and the integrals become easy.

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