Question

Let the curve \(y = y(x)\) be the solution of the differential equation, \(\frac{dy}{dx} = 2(x+1)\). If the numerical value of area bounded by the curve \(y = y(x)\) and x-axis is \(\frac{4\sqrt8}{3}\), then the value of \(y(1)\) is equal to ____________.

Hint:

Integrate to get \(y=(x+1)^2+c\); the curve meets the x-axis only if \(c \lt 0\), and the enclosed area is \(\frac{4}{3}(\sqrt{-c})^3\).

Answer 1

Correct Answer: 2

\(\frac{dy}{dx} = 2(x+1) \Rightarrow y = (x+1)^2 + c\). For a bounded area with the x-axis the parabola must cut the axis, so \(c \lt 0\). Let \(s = \sqrt{-c}\); the roots are \(x = -1 \pm s\).
Area \(= \int_{-1-s}^{-1+s} \left(s^2 - (x+1)^2\right)dx = \left[s^2 u - \frac{u^3}{3}\right]_{-s}^{s} = 2\left(s^3 - \frac{s^3}{3}\right) = \frac{4s^3}{3}\) (with \(u = x+1\)).
\(\frac{4s^3}{3} = \frac{4\sqrt8}{3} \Rightarrow s^3 = 8^{1/2} = 2^{3/2} \Rightarrow s = \sqrt2\), so \(c = -2\).
\(y(1) = (1+1)^2 - 2 = 2\).
Answer: 2