Consider the curve which is represented by the differential equation $\frac{dy}{dx}=1+x+y+ xy$. 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 $(-1,\frac{1}{\sqrt{e}}-1)$
(D) Its equation is $xy(x + 1)(y − \frac{1}{\sqrt{e}} + 1) = 0$

Choose the correct answer from the options given below:

(C) only

The correct answer is Option (4) → (C) only

(A) it is a straight line. (False)
(B) it is a parabola. (False)
(C) it also passes through the point $(-1,\frac{1}{\sqrt{e}},-1)$ (True)
(D) Its equation is $xy(x + 1)(y − \frac{1}{\sqrt{e}} + 1) = 0$ (False)

Given differential equation:

$\frac{dy}{dx}=1+x+y+xy=(1+x)(1+y)$

Separate variables:

$\frac{1}{1+y}\,dy=(1+x)\,dx$

Integrate:

$\ln(1+y)=x+\frac{x^{2}}{2}+C$

Use initial condition $(0,0)$:

$0=\ln(1)=0+\frac{0}{2}+C$

$C=0$

Thus the solution is:

$\ln(1+y)=x+\frac{x^{2}}{2}$

$1+y=e^{\,x+\frac{x^{2}}{2}}$

$y=e^{\,x+\frac{x^{2}}{2}}-1$

Now check each option:

(A) It is a straight line.

False, because $y=e^{\,x+\frac{x^{2}}{2}}-1$ is not linear.

(B) It is a parabola.

False, the solution is exponential, not polynomial.

(C) It passes through $\left(-1,\frac{1}{\sqrt{e}}-1\right)$.

At $x=-1$:

$y=e^{-1+\frac{1}{2}}-1=e^{-1/2}-1=\frac1{\sqrt e}-1$

True.

(D) Equation: $xy(x+1)(y-\frac1{\sqrt e}+1)=0$

False, does not match the actual implicit solution

$\ln(1+y)=x+\frac{x^{2}}{2}$.

Correct option: (C)