If dy/dx = (2ˣy + 2ʸ.2ˣ)/(2ˣ + 2ˣ⁺ʸlog ₑ2), y(0) = 0, then for y = 1, the value of x lies in the interval :

Solution

Maths sir · 4 steps

Cancel 2ˣ first, then notice the denominator is exactly the derivative of the numerator in y.

1. Cancel the common factor. The numerator is 2ˣ(y + 2ʸ) and the denominator is 2ˣ(1 + 2ʸ ln 2). Cancelling 2ˣ leaves dy/dx = (y + 2ʸ)/(1 + 2ʸ ln 2), which has no x in it at all.
2. Separate the variables. Move all the y terms to one side. Since d/dy(y + 2ʸ) = 1 + 2ʸ ln 2, the left side is of the form f'(y)/f(y), and that integrates to a logarithm.

   ∫ (1 + 2ʸ ln 2)/(y + 2ʸ) dy = ∫ dx ⇒ ln(y + 2ʸ) = x + C

3. Use the initial condition. At x = 0, y = 0, so ln(0 + 1) = 0 + C, which gives C = 0. The solution is x = ln(y + 2ʸ).
4. So. At y = 1, x = ln(1 + 2) = ln 3 ≈ 1.099. Since e < 3 < e², x lies between 1 and 2, so the answer is (a).

The idea to keep. Before integrating, check whether the top or bottom is the derivative of the other. If it is, ∫ f'/f = ln|f| finishes the problem in one line.

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