Let y=y(x) be the solution of the differential equation (x²+4)² d y+(2 x³ y+8 x y-2) d x=0. If y(0)=0, then y(2) is equal to

Solution

Maths sir · 4 steps

Factor the y terms as 2xy(x²+4) and you have a linear equation with integrating factor x²+4.

1. Make it linear. The dx terms group as 2x³y + 8xy = 2xy(x²+4). Divide the whole equation by (x²+4)² dx to get dy/dx + 2x/(x²+4) y = 2/(x²+4)².
2. Integrating factor. Here P = 2x/(x²+4), so ∫ P dx = ln(x²+4) and the integrating factor is x²+4. Multiplying by it turns the left side into the derivative of y(x²+4).

   d/dx[y (x²+4)] = 2/(x²+4) ⇒ y (x²+4) = tan⁻¹ x/2 + C

3. Use y(0) = 0. At x = 0 the left side is 0 and tan⁻¹ 0 = 0, so C = 0. That gives y = 1/(x²+4)tan⁻¹ x/2.
4. So. y(2) = 1/8 tan⁻¹ 1 = 1/8 × π/4 = π/32, which is option (d).

The idea to keep. Before reaching for any method, factor the terms with y. When the coefficient becomes f'(x)/f(x), the integrating factor is just f(x) and the left side collapses at once.

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