If ydy/dx = x[ y²/x² + (φ ( y²/x² ))/(φ '( y²/x² )) ], x > 0, φ > 0, and y(1) = -1, then φ ( y²/4 ) is equal to :

Solution

Maths sir · 4 steps

The ratio y²/x² appears everywhere, so call it u; the equation then separates neatly.

1. Substitute. Put u = y²/x², so y² = ux². Differentiating, 2ydy/dx = x²du/dx + 2ux, which gives ydy/dx = x²/2du/dx + ux.
2. Cancel and separate. The right side of the given equation is ux + xφ(u)/φ'(u). The ux terms cancel, leaving x²/2du/dx = xφ(u)/φ'(u).

   ∫ φ'(u)/φ(u) du = ∫ 2/x dx ⇒ ln φ(u) = 2 ln x + c ⇒ φ(u) = Cx²

3. Use y(1) = -1. At x = 1, y = -1 we get u = 1, so φ(1) = C. Hence φ(y²/x²) = x² φ(1) along the solution curve.
4. So. φ(y²/4) is this relation at x = 2, which gives 2² φ(1) = 4φ(1). The answer is (b).

The idea to keep. When one ratio such as y²/x² keeps repeating, make it your new variable. Here φ'/φ integrates straight to ln φ.

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