Consider the differential equation \(\displaystyle\frac{dy}{dx} = y^2\).
(A) Its general solution is \(y = -\displaystyle\frac{1}{x + C}\), where C is an arbitrary constant.
(B) \(y = 0\) is a solution of the equation.
(C) The solution satisfying \(y(0) = 1\) is \(y = \displaystyle\frac{1}{1 - x}\).
(D) The solution satisfying \(y(0) = 1\) has \(y\left(\displaystyle\frac{1}{2}\right) = \displaystyle\frac{3}{2}\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
For \(y \ne 0\) the variables separate:
\( \displaystyle \frac{dy}{y^2} = dx \)
Integrating both sides:
\( \displaystyle -\frac{1}{y} = x + C \)
\( \displaystyle y = -\frac{1}{x + C} \)
With y(0) = 1: \(-1 = C\), so
\( \displaystyle y = -\frac{1}{x - 1} = \frac{1}{1 - x} \)
At \(x = \frac{1}{2}\):
\( \displaystyle y = \frac{1}{1/2} = 2 \)
So the general solution is \(y = -\frac{1}{x + C}\), the solution with y(0) = 1 is \(\frac{1}{1 - x}\), and its value at \(\frac{1}{2}\) is 2.
Checking the statements
(A) General solution \(y = -\frac{1}{x + C}\) — Correct. It comes from \(-\frac{1}{y} = x + C\).
(B) \(y = 0\) is a solution — Correct. Both sides are 0: \(\frac{dy}{dx} = 0\) and \(y^2 = 0\).
(C) The solution with y(0) = 1 is \(\frac{1}{1 - x}\) — Correct. The condition gives C = −1.
(D) \(y\left(\frac{1}{2}\right) = \frac{3}{2}\) — Incorrect. \(\frac{1}{1 - 1/2} = 2\).