If the curve, y = y(x) represented by the solution of the differential equation (2xy² - y)dx + xdy = 0, passes through the intersection of the lines, 2x - 3y = 1 and 3x + 2y = 8, then |y(1)| is equal to _________.

Solution

Maths sir · 4 steps

See x dy - y dx sitting there? Divide by y² and it becomes the exact differential of x/y.

1. Regroup the terms. Expand to 2xy² dx - y dx + x dy = 0 and move the y² term across: y dx - x dy = 2xy² dx. Dividing by y² turns the left side into d(x/y).

   (y dx - x dy)/y² = 2x dx ⇒ d(x/y) = 2x dx ⇒ x/y = x² + C

2. Find the point. Solve 2x - 3y = 1 and 3x + 2y = 8. Doubling the first and tripling the second gives 4x - 6y = 2 and 9x + 6y = 24. Adding them gives 13x = 26, so x = 2 and y = 1.
3. Fix the constant. Put (2, 1) into x/y = x² + C: 2 = 4 + C, so C = -2. The curve is x/y = x² - 2.
4. So. At x = 1: 1/y = 1 - 2 = -1, so y(1) = -1 and |y(1)| = 1.

The common slip: 0.2. Writing d(x/y) as (x dy - y dx)/y² flips the sign and gives x/y = 6 - x², so |y(1)| = 0.2 instead of 1.

The idea to keep. Learn to spot x dy - y dx and y dx - x dy. Divide by x² or y² and they become d(y/x) or d(x/y), and the equation integrates in one line.

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