Let y=y(x) be the solution of the differential equation
(d y)/(d x)=(4 y³+2 y x²)/(3 x y²+x³), y(1)=1.
If for some n ∈ ℕ, y(2) ∈[n-1, n), then n is equal to _____________.

Solution

Maths sir · 5 steps

Every term is degree 3, so put y = vx. The numerator then becomes the exact derivative of the denominator.

1. Substitute y = vx. Each term on top and bottom has degree 3, so the equation is homogeneous. With y = vx we get dy/dx = v + xdv/dx, and the right side becomes (4v³ + 2v)/(3v² + 1).
2. Separate variables. Move v to the right: xdv/dx = (4v³ + 2v - 3v³ - v)/(3v²+1) = (v³ + v)/(3v² + 1). The top 3v² + 1 is the derivative of v³ + v, so both sides integrate to logarithms.

   ∫ (3v²+1)/(v³+v) dv = ∫ dx/x ⇒ ln(v³+v) = ln x + C

3. Use y(1) = 1. At x = 1, v = 1, so C = ln 2 and v³ + v = 2x. Put back v = y/x and multiply by x³: y³ + x² y = 2x⁴.
4. Locate y(2). At x = 2: y³ + 4y = 32. The left side increases with y. It equals 16 at y = 2 and 39 at y = 3, so the root lies between them, near y ≈ 2.76.
5. So. y(2) ∈ [2, 3), which is [n-1, n) with n = 3.

The common slip: 2. Reporting the integer part of y(2) gives 2, but the interval is [n-1, n), so n is one more than that.

The idea to keep. When all terms share one degree, use y = vx. After separating, check whether the numerator is the derivative of the denominator. If it is, the integral is a log and you get the curve in one line.

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