If the solution y=y(x) of the differential equation (x⁴+2 x³+3 x²+2 x+2) d y-(2 x²+2 x+3) d x=0 satisfies y(-1)=-π/4, then y(0) is equal to :

Solution

Maths sir · 4 steps

Factor the quartic into two quadratics, then split the numerator to match them: two arctans fall out.

1. Separate the variables. Rearranging gives dy/dx = (2x²+2x+3)/(x⁴+2x³+3x²+2x+2). The whole job is to integrate this fraction.
2. Factor the denominator. The quartic factors as (x²+1)(x²+2x+2). Multiplying out gives x⁴+2x³+3x²+2x+2. The numerator is exactly the sum of these two factors: 2x²+2x+3 = (x²+1)+(x²+2x+2).

   dy/dx = 1/(x²+2x+2) + 1/(x²+1) = 1/((x+1)²+1) + 1/(x²+1)

3. Integrate. This gives y = tan⁻¹(x+1) + tan⁻¹ x + C. At x=-1 we get 0 + (-π/4) + C = -π/4, so C = 0.
4. So. y(0) = tan⁻¹ 1 + tan⁻¹ 0 = π/4, which is option (d).

The idea to keep. When a big rational function shows up, try to factor the denominator and write the numerator as a sum of those factors. The fraction then splits into standard integrals with no partial-fraction algebra.

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