Let f be a differentiable function such that 2(x+2)² f(x)-3(x+2)²=10 ∫₀ˣ(t+2) f(t) d t, x ≥ 0. Then f(2) is equal to ________ .

Solution

Maths sir · 4 steps

Differentiate once to kill the integral, and put x = 0 in the original equation to get f(0).

1. Starting value. At x = 0 the integral is 0, so 2(4)f(0) - 3(4) = 0. This gives f(0) = 3/2.
2. Differentiate. Differentiating both sides gives 4(x+2)f + 2(x+2)² f' - 6(x+2) = 10(x+2)f. Since x + 2 ≠ 0 for x ≥ 0, divide by x+2 to get 2(x+2)f' = 6f + 6.

   df/(f+1) = (3 dx)/(x+2) ⇒ f + 1 = k(x+2)³

3. Find the constant. Put x = 0: 3/2 + 1 = 8k, so k = 5/16. Then f(x) = 5/16(x+2)³ - 1.
4. So. f(2) = 5/16 × 64 - 1 = 20 - 1 = 19.

The common slip: 20. Reporting f(2) + 1 instead of f(2) gives 20. The -1 must come off at the end.

The idea to keep. When f sits inside an integral from a fixed point to x, differentiate to get a differential equation. Then put x equal to the lower limit in the original equation to get the initial condition.

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