Question

The general solution of the differential equation \( \frac{dy}{dx}=x^{2}y(x+y+xy+1) \) is \( y = \)

• A. \( A \cdot e^{\frac{x^{3}}{3}} \cdot e^{\frac{x^{4}}{4}}(1+y) \)
• B. \( (y+1)Ae^{\dots} \)
• C. \( (y+1) + A \cdot e^{\frac{x^{3}}{3}} \cdot e^{\frac{x^{4}}{4}} \)
• D. \( \frac{A \cdot e^{\frac{x^{3}}{3}}(y+1)}{e^{\frac{x^{4}}{4}}} \)

Hint:

Always check for grouping patterns in multivariate polynomials. Factoring out \( (x+1) \) turns a complicated-looking differential string into a standard separable variable equation instantly.

Answer 1

The Correct Option is A

Concept: We first look for common algebraic factors inside the parentheses to group terms, allowing us to solve the differential equation using the method of separation of variables.

Step 1: Factorizing the expression inside the parentheses.
\[ x + y + xy + 1 = (x + 1) + y(x + 1) = (x + 1)(y + 1) \] Substituting this back into the differential equation: \[ \frac{dy}{dx} = x^2 y (x + 1)(y + 1) \implies \frac{dy}{dx} = (x^3 + x^2) y(y + 1) \]

Step 2: Separating variables and integrating.
Group all \( y \) terms on the left side and all \( x \) terms on the right side: \[ \frac{1}{y(y + 1)} dy = (x^3 + x^2) dx \] Using partial fractions on the left side: \[ \int \left( \frac{1}{y} - \frac{1}{y+1} \right) dy = \int (x^3 + x^2) dx \] \[ \log|y| - \log|y+1| = \frac{x^4}{4} + \frac{x^3}{3} + C \implies \log\left| \frac{y}{y+1} \right| = \frac{x^4}{4} + \frac{x^3}{3} + C \]

Step 3: Converting from logarithmic to exponential form.
Taking the exponential on both sides: \[ \frac{y}{y+1} = e^C \cdot e^{\frac{x^3}{3}} \cdot e^{\frac{x^4}{4}} \implies y = A \cdot e^{\frac{x^3}{3}} \cdot e^{\frac{x^4}{4}}(y+1) \] This precisely tracks the structural layout of Option (A).