Let y=y(x) be the solution of the differential equation (x²-3y²)dx+3xy dy=0,y(1)=1. Then 6y²(e) is equal to

Solution

Maths sir · 4 steps

Every term has degree 2, so put y = vx: the equation separates in one line.

1. Spot homogeneous. Rewrite it as dy/dx = (3y² - x²)/3xy. Every term in the top and bottom has degree 2, so substitute y = vx, which gives dy/dx = v + xdv/dx.
2. Separate variables. Then v + xdv/dx = (3v² - 1)/3v, so xdv/dx = (3v² - 1 - 3v²)/3v = -1/3v. This separates as 3v dv = -dx/x.

   3/2v² = -ln x + C

3. Use y(1) = 1. At x = 1, v = y/x = 1 and ln 1 = 0, so C = 3/2. Multiply through by 2/3 and put v = y/x back to get y² = x²(1 - 2/3 ln x).
4. So. At x = e, ln e = 1, so y²(e) = e² × 1/3. Then 6y²(e) = 2e², which is option (d).

The idea to keep. When every term of dy/dx has the same degree in x and y, put y = vx. The v terms usually cancel, and what remains separates.

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