Let the solution curve of the differential equation x d y=(√(x²+y²)+y) d x, x>0, intersect the line x=1 at y=0 and the line x=2 at y=α. Then the value of α is :

Solution

Maths sir · 5 steps

Every term here has degree 1 in x and y, so put y = vx and the variables separate.

1. Spot homogeneous. Write it as dy/dx = (√(x²+y²)+y)/x. The numerator and the denominator are both degree 1, so substitute y = vx, which gives dy/dx = v + xdv/dx.
2. Substitute. Since x > 0, √(x²+v²x²) = x√(1+v²). So v + xdv/dx = √(1+v²) + v. The v cancels, leaving xdv/dx = √(1+v²).

   ∫ dv/√(1+v²) = ∫ dx/x ⇒ ln(v + √(1+v²)) = ln x + C

3. Find the constant. At x = 1, y = 0, so v = 0. The left side is ln 1 = 0 and the right side is 0 + C, so C = 0. The curve is v + √(1+v²) = x.
4. Set x to 2. √(1+v²) = 2 - v. Squaring gives 1 + v² = 4 - 4v + v², so v = 3/4. This value keeps 2 - v positive, so it is valid.
5. So. α = y = vx = 3/4 × 2 = 3/2, which is option (b).

The idea to keep. When every term has the same degree, y = vx separates the variables. Always multiply back by x at the end: the question asks for y, not v.

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