Consider the differential equation \(\displaystyle\frac{dy}{dx} = 1 + y^2\) and the following statements:
(A) It is of order 1 and degree 1.
(B) Its general solution is \(y = \tan (x + C)\), where C is an arbitrary constant.
(C) The solution satisfying \(y(0) = 1\) has \(y\left(\displaystyle\frac{\pi}{12}\right) = \sqrt{3}\).
(D) The solution satisfying \(y(0) = 0\) is \(y = \tan^{-1} x\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (C) only
Only the first derivative appears, with power 1. Separating the variables:
\( \displaystyle \frac{dy}{1 + y^2} = dx \)
Integrating both sides:
\( \tan^{-1} y = x + C \)
\( y = \tan (x + C) \)
For \(y(0) = 1\): \(\tan C = 1\), so \(C = \frac{\pi}{4}\) and
\( \displaystyle y\left(\frac{\pi}{12}\right) = \tan\left(\frac{\pi}{12} + \frac{\pi}{4}\right) = \tan \frac{\pi}{3} = \sqrt{3} \)
For \(y(0) = 0\): \(\tan C = 0\), so \(C = 0\) and
\( y = \tan x \)
So the equation has order 1 and degree 1, its general solution is y = tan(x + C), the solution with y(0) = 1 gives \(y\left(\frac{\pi}{12}\right) = \sqrt{3}\), and the solution with y(0) = 0 is y = tan x.
Checking the statements
(A) It is of order 1 and degree 1. — Correct. Only \(\frac{dy}{dx}\) appears, with power 1.
(B) Its general solution is \(y = \tan (x + C)\), where C is an arbitrary constant. — Correct. It comes from \(\tan^{-1} y = x + C\).
(C) The solution satisfying \(y(0) = 1\) has \(y\left(\frac{\pi}{12}\right) = \sqrt{3}\). — Correct. Here \(C = \frac{\pi}{4}\), and \(\tan \frac{\pi}{3} = \sqrt{3}\).
(D) The solution satisfying \(y(0) = 0\) is \(y = \tan^{-1} x\). — Incorrect. Here C = 0, so the solution is \(y = \tan x\).