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The general solution of the differential equation dydx=1+x+y+xy\dfrac{dy}{dx} = 1+x+y+xy is: (where CC is an arbitrary constant)

(A) 1+y=x+x22+C1+y = x+\dfrac{x^2}{2}+C

(B) log⁡e∣1+y∣2=x2+2x+C\log_e|1+y|^2 = x^2+2x+C

(C) log⁡e∣1−y∣=x−x22+C\log_e|1-y| = x-\dfrac{x^2}{2}+C

(D) log⁡e∣1+y∣=x+x22+C\log_e|1+y| = x+\dfrac{x^2}{2}+C

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

Factorising, dydx=(1+x)(1+y)\dfrac{dy}{dx}=(1+x)(1+y), so dy1+y=(1+x)dx\dfrac{dy}{1+y}=(1+x)dx and log⁡e∣1+y∣=x+x22+C\log_e|1+y|=x+\dfrac{x^2}{2}+C, which is (D). Multiplying throughout by 2 gives log⁡e∣1+y∣2=x2+2x+C\log_e|1+y|^2=x^2+2x+C, which is (B).

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