The general solution of the differential equation \( \displaystyle\frac{dy}{dx}=1+x+y+xy \) is : (where \( C \) is an arbitrary constant)
(A) \( 1+y=x+\displaystyle\frac{x^2}{2}+C \)
(B) \( \log_e|1+y|^2=x^2+2x+C \)
(C) \( \log_e|1-y|=x-\displaystyle\frac{x^2}{2}+C \)
(D) \( \log_e|1+y|=x+\displaystyle\frac{x^2}{2}+C \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (D) only
The right side factorises:
\( \displaystyle \frac{dy}{dx}=(1+x)(1+y) \)
Separating the variables,
\( \displaystyle \frac{dy}{1+y}=(1+x)\,dx \)
Integrating both sides,
\( \displaystyle \log_e|1+y|=x+\frac{x^2}{2}+C \)
Multiplying by 2 and writing the constant 2C as C,
\( \displaystyle \log_e|1+y|^2=x^2+2x+C \)
So the general solution is \( \log_e|1+y|=x+\frac{x^2}{2}+C \), which can also be written as \( \log_e|1+y|^2=x^2+2x+C \).
Checking the statements
(A) \( 1+y=x+\displaystyle\frac{x^2}{2}+C \) — Incorrect. The left side must be \( \log_e|1+y| \), not 1 + y.
(B) \( \log_e|1+y|^2=x^2+2x+C \) — Correct. It is twice the solution, with 2C written as C.
(C) \( \log_e|1-y|=x-\displaystyle\frac{x^2}{2}+C \) — Incorrect. The solution has \( \log_e|1+y| \) and \( +\frac{x^2}{2} \); both signs are wrong here.
(D) \( \log_e|1+y|=x+\displaystyle\frac{x^2}{2}+C \) — Correct. It is exactly the solution \( \log_e|1+y|=x+\frac{x^2}{2}+C \).