If y = y(x) is the solution of the differential equation, eʸ( dy/dx - 1 ) = eˣ such that y(0) = 0, then y(1) is equal to:

Solution

Maths sir · 3 steps

Divide by eˣ, and the equation becomes separable in one new variable, t = y - x.

1. Spot the substitution. Divide both sides by eˣ to get eʸ⁻ˣ(dy/dx - 1) = 1. Put t = y - x. Then dt/dx = dy/dx - 1, so the bracket is exactly dt/dx.

   eᵗ dt = dx ⇒ eᵗ = x + C

2. Use y(0) = 0. At x = 0, t = y - x = 0, so e⁰ = 0 + C gives C = 1. The solution is eʸ⁻ˣ = x + 1, that is, y = x + logₑ(x+1).
3. So. y(1) = 1 + logₑ 2, which is option (c).

The idea to keep. When dy/dx - 1 or dy/dx + 1 appears next to eʸ and eˣ, try t = y ∓ x. The equation usually separates straight away.

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