Let y = y(x) be the solution of the differential equation ( (x + 2)e^(( (y + 1)/(x + 2) )) + (y + 1) )dx = (x + 2)dy, y(1) = 1. If the domain of y = y(x) is an open interval (α, β), then | α + β| is equal to ______________. (Round off to the nearest integer)

Solution

Maths sir · 5 steps

Shift to X = x+2, Y = y+1. The equation becomes homogeneous, so put Y = vX.

1. Shift the origin. Let X = x + 2 and Y = y + 1, so dX = dx and dY = dy. Dividing by X dX gives dY/dX = e^(Y/X) + Y/X, a homogeneous equation.
2. Separate variables. Put Y = vX, so dY/dX = v + Xdv/dX. The v cancels, leaving Xdv/dX = eᵛ, that is e⁻ᵛ dv = dX/X. Integrating gives -e⁻ᵛ = ln|X| + C.
3. Use y(1) = 1. At x = 1 we have X = 3, Y = 2, v = 2/3. So C = -ln 3 - e^(-2/3), and the solution is shown below.

   e^(-(y+1)/(x+2)) = ln 3 + e^(-2/3) - ln(x+2)

4. Find the domain. The left side is always positive, so ln(x+2) < ln 3 + e^(-2/3), which gives x + 2 < 3e^(e^(-2/3)). The solution passes through X = 3 > 0 and cannot cross X = 0, so x + 2 > 0. The domain is (-2, 3e^(e^(-2/3)) - 2).
5. So. α + β = 3e^(e^(-2/3)) - 4. Since e^(-2/3) ≈ 0.513 and e^(0.513) ≈ 1.671, this is 5.013 - 4 ≈ 1.01. So |α + β| rounds to 1.

The common slip: 4. Writing the domain as |x+2| < 3e^(e^(-2/3)) ignores that the curve cannot cross x = -2. That makes the interval symmetric about -2, so α + β = -4 and the answer comes out as 4.

The idea to keep. When you see (x+a) and (y+b) together, shift the origin first. For the domain of the solution, ask what keeps every term defined and every positive quantity positive, starting from the given point.

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