If the solution curve f(x, y)=0 of the differential equation
(1+log ₑ x) (d x)/(d y)-x log ₑ x=eʸ, x > 0,
passes through the points (1,0) and (α, 2), then α^α is equal to :

Solution

Maths sir · 4 steps

Notice 1+ln x is the derivative of x ln x, so put t = x ln x and the equation turns linear.

1. Substitute. Let t = x ln x. Then dt/dx = 1 + ln x, so dt/dy = (1+ln x)dx/dy. The equation becomes dt/dy - t = eʸ, which is linear in t.

   d/dy(t e⁻ʸ) = eʸ· e⁻ʸ = 1 ⇒ t e⁻ʸ = y + C

2. Use the point. At (1, 0), t = 1 · ln 1 = 0 and y = 0, so C = 0. The solution curve is x ln x = y eʸ.
3. Put y = 2. At (α, 2) we get α ln α = 2e². The left side is ln(α^α).
4. So. ln(α^α) = 2e², which gives α^α = e^(2e²). The answer is (d).

The idea to keep. When a messy coefficient is the derivative of another term in the equation, make that term the new variable. Here x ln x became t and the equation turned into a standard linear one.

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