If y=y(x) is the solution curve of the differential equation (x²-4) d y-(y²-3 y) d x=0, x>2, y(4)=3/2 and the slope of the curve is never zero, then the value of y(10) equals :

Solution

Maths sir · 5 steps

Separate the variables, use partial fractions on both sides, and let the slope condition fix the sign.

1. Separate. Rewrite the equation as dy/(y²-3y) = dx/(x²-4). On the left, 1/y(y-3) = 1/3(1/(y-3)-1/y). On the right, use ∫ dx/(x²-a²) = 1/2a ln|(x-a)/(x+a)| with a = 2.

   1/3 ln|(y-3)/y| = 1/4 ln((x-2)/(x+2)) + C

2. Use y(4) = 3/2. At y = 3/2, (y-3)/y = -1, so the left side is ln 1 = 0. The right side is 1/4 ln 1/3 + C, so C = 1/4 ln 3. This gives |(y-3)/y| = (3(x-2)/(x+2))^(3/4).
3. Put x = 10. (3 × 8)/12 = 2, so |(y-3)/y| = 2^(3/4) = 8^(1/4).
4. Fix the sign. The slope is (y²-3y)/(x²-4), which is zero only when y = 0 or y = 3. The slope is never zero, so the curve never reaches these values. Since it starts at y = 3/2, it stays in 0 < y < 3, where (y-3)/y is negative.
5. So. (3-y)/y = 8^(1/4), which gives 3/y = 1 + 8^(1/4) and y(10) = 3/(1+8^(1/4)), option (a).

The idea to keep. When a modulus appears after integrating, do not drop it blindly. Use the given conditions (here, slope never zero) to find which interval y stays in, and that decides the sign.

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