JEE Main 2022 · 29 June · morning · Single correct · +4 / −1

Let the solution curve of the differential equation
xdy/dx - y = √(y² + 16x²), y(1) = 3 be y = y(x). Then y(2) is equal to:

  1. (a) 15
  2. (b) 11
  3. (c) 13
  4. (d) 17
Answer and solution

Answer (a) 15

Solution

Maths sir · 5 steps

Divide by x and you see everything depends on y/x, so put y = vx and separate.

  1. Spot homogeneous form. Near x = 1 we have x > 0, so dividing by x gives dy/dx = y/x + √((y/x)² + 16). The right side depends only on y/x, so put y = vx, which gives dy/dx = v + xdv/dx.
  2. Separate variables. The v on both sides cancels, leaving xdv/dx = √(v² + 16). Integrate both sides with the standard result ∫ dv/√(v² + a²) = ln|v + √(v² + a²)|.

    ln(v + √(v² + 16)) = ln x + ln C ⇒ v + √(v² + 16) = Cx

  3. Use the condition. At x = 1, y = 3, so v = 3. Then 3 + √(9 + 16) = 3 + 5 = 8 = C. The curve is v + √(v² + 16) = 8x.
  4. Evaluate at two. At x = 2, v + √(v² + 16) = 16, so √(v² + 16) = 16 - v. Squaring gives v² + 16 = 256 - 32v + v², so 32v = 240 and v = 7.5.
  5. So. y(2) = vx = 7.5 × 2 = 15, which is option (a).

The idea to keep. If the right side can be written using only y/x, put y = vx. The v terms cancel and the variables separate. Find the constant from the given point before you plug in the new x.

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

Study Differential Equations

Asked about 200 times in JEE Main since 2020. This question is from 2022.