JEE Main 2020 · 9 January · evening · Single correct · +4 / −1
If dy/dx = xy/(x² + y²); y(1) = 1; then a value of x satisfying y(x) = e is :
- (a) √2 e
- (b) 1/2√3 e
- (c) e/√2
- (d) √3 e
Answer and solution
Answer (d) √3 e
Solution
Maths sir · 5 steps
Every term on the right has degree 2, so put y = vx and the variables separate.
- Substitute y = vx. Numerator and denominator are both of degree 2, so the equation is homogeneous. With y = vx we get dy/dx = v + xdv/dx, and the right side becomes v/(1+v²).
- Separate. xdv/dx = v/(1+v²) - v = -v³/(1+v²). This rearranges to (1/v³ + 1/v)dv = -dx/x.
- Integrate. Integrating gives -1/2v² + ln v = -ln x + C. Put v = y/x, so ln v + ln x = ln y, and the solution becomes the equation below.
ln y - x²/2y² = C
- Find C. At y(1) = 1: 0 - 1/2 = C, so C = -1/2. Now put y = e: 1 - x²/2e² = -1/2, which gives x²/2e² = 3/2.
- So. x² = 3e², so x = √3 e. That is option (d).
The idea to keep. If every term in dy/dx has the same degree, put y = vx and separate. Fix the constant from the initial condition before you substitute the new point.
Answer from the NTA final answer key. Solution by PrepWiser.