Consider the curve which is represented by the differential equation \(\displaystyle\frac{dy}{dx} = 1 + x + y + xy\). If it passes through the point (0, 0), then which of the following is/are true?
(A) it is a straight line.
(B) it is a parabola.
(C) it also passes through the point \(\left(-1, \displaystyle\frac{1}{\sqrt{e}} - 1\right)\)
(D) Its equation is \(xy(x + 1)\left(y - \displaystyle\frac{1}{\sqrt{e}} + 1\right) = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (C) only
The right side factorises, so the variables separate:
\( \displaystyle \frac{dy}{dx} = (1 + x) + y(1 + x) = (1 + x)(1 + y) \)
\( \displaystyle \int \frac{dy}{1 + y} = \int (1 + x)\,dx \)
\( \displaystyle \log|1 + y| = x + \frac{x^2}{2} + C \)
At (0, 0): \(\log 1 = 0 + C\), so \(C = 0\). Near (0, 0), \(1 + y \gt 0\), so
\( \displaystyle y = e^{x + \frac{x^2}{2}} - 1 \)
So the curve is \(y = e^{x + \frac{x^2}{2}} - 1\).
Checking the statements
(A) it is a straight line. — Incorrect. The curve is exponential, not \(y = mx + c\).
(B) it is a parabola. — Incorrect. y is not a polynomial of degree 2 in x.
(C) it also passes through the point \(\left(-1, \frac{1}{\sqrt{e}} - 1\right)\) — Correct. At \(x = -1\):
\( \displaystyle y = e^{-1 + \frac{1}{2}} - 1 = e^{-\frac{1}{2}} - 1 = \frac{1}{\sqrt{e}} - 1 \)
(D) Its equation is \(xy(x + 1)\left(y - \frac{1}{\sqrt{e}} + 1\right) = 0\) — Incorrect. It is the four lines \(x = 0\), \(y = 0\), \(x = -1\), \(y = \frac{1}{\sqrt{e}} - 1\); the curve's point \(\left(1, e^{3/2} - 1\right)\) is on none of them.