Consider the differential equation \(\displaystyle\frac{dy}{dx} = 4x\sqrt{y}\), y > 0.
(A) It is of order 1 and degree 2.
(B) Its general solution can be written as \(\sqrt{y} = x^2 + C\), where C is an arbitrary constant.
(C) The solution satisfying y(0) = 1 is \(y = \left(x^2 + 1\right)^2\).
(D) The solution satisfying y(0) = 4 has y(1) = 9.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B), (C) and (D) only
The only derivative is \(\frac{dy}{dx}\), with power 1, so the order is 1 and the degree is 1. The square root is on y, not on a derivative.
The variables separate:
\( \displaystyle \frac{dy}{2\sqrt{y}} = 2x\,dx \)
Integrating both sides:
\(\sqrt{y} = x^2 + C\)
With y(0) = 1: \(C = 1\), so
\(y = \left(x^2 + 1\right)^2\)
With y(0) = 4: \(C = 2\), so
\(y = \left(x^2 + 2\right)^2\)
\(y(1) = 3^2 = 9\)
So the order and degree are both 1, the general solution is \(\sqrt{y} = x^2 + C\), and the two particular solutions are \(\left(x^2 + 1\right)^2\) and \(\left(x^2 + 2\right)^2\).
Checking the statements
(A) Order 1 and degree 2 — Incorrect. \(\frac{dy}{dx}\) appears with power 1, so the degree is 1.
(B) General solution \(\sqrt{y} = x^2 + C\) — Correct. It comes from integrating \(\frac{dy}{2\sqrt{y}} = 2x\,dx\).
(C) The solution with y(0) = 1 is \(\left(x^2 + 1\right)^2\) — Correct. The condition gives C = 1.
(D) The solution with y(0) = 4 has y(1) = 9 — Correct. C = 2, so \(y(1) = (1 + 2)^2 = 9\).