If y = y(x) is the solution of the differential equation \(\displaystyle\frac{dy}{dx} = \frac{3x^2 + 1}{2y + 1}\) with y(0) = 1 and y > 0, then y(2) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
The variables separate:
\((2y + 1)\,dy = (3x^2 + 1)\,dx\)
Integrating both sides:
\(y^2 + y = x^3 + x + C\)
With y(0) = 1:
\(1 + 1 = 0 + C\)
\(C = 2\)
At x = 2:
\(y^2 + y = 8 + 2 + 2 = 12\)
\(y^2 + y - 12 = 0\)
\((y + 4)(y - 3) = 0\)
Since y > 0:
\(y(2) = 3\)