Let slope of the tangent line to a curve at any point P(x, y) be given by (xy² + y)/x. If the curve intersects the line x + 2y = 4 at x = -2, then the value of y, for which the point (3, y) lies on the curve, is :

Solution

Maths sir · 4 steps

Multiply out by x and spot x dy - y dx: divided by y², it is an exact differential.

1. Rearrange. From dy/dx = (xy² + y)/x we get x dy = xy² dx + y dx, so x dy - y dx = xy² dx. Divide both sides by y².

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

2. Find the point. At x = -2 the line x + 2y = 4 gives 2y = 6, so y = 3. The curve passes through (-2, 3).
3. Find C. Put in (-2, 3): --2/3 = 4/2 + C, so 2/3 = 2 + C and C = -4/3.
4. So. At x = 3: -3/y = 9/2 - 4/3 = 19/6, so y = -18/19. The answer is (a).

The idea to keep. When a first-order equation has x dy - y dx in it, divide by y² or x². It becomes -d(x/y) or d(y/x), and you can integrate in one line.

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