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

Step 1: Factor the bracket.
Group terms inside: \[ x+y+xy+1=(x+1)+y(x+1)=(x+1)(y+1) \]
Step 2: Rewrite the equation.
\[ \frac{dy}{dx}=x^2y(x+1)(y+1)=(x^3+x^2)\,y(y+1) \]
Step 3: Separate the variables.
Put all $y$ on one side and all $x$ on the other: \[ \frac{dy}{y(y+1)}=(x^3+x^2)\,dx \]
Step 4: Integrate using partial fractions.
Since $\frac{1}{y(y+1)}=\frac{1}{y}-\frac{1}{y+1}$, \[ \log|y|-\log|y+1|=\frac{x^4}{4}+\frac{x^3}{3}+C \]
Step 5: Combine the logs.
\[ \log\left|\frac{y}{y+1}\right|=\frac{x^4}{4}+\frac{x^3}{3}+C \]
Step 6: Take exponentials.
\[ \frac{y}{y+1}=A\,e^{x^3/3}\,e^{x^4/4}\implies y=A\,e^{x^3/3}\,e^{x^4/4}(1+y) \] \[ \boxed{y=A\,e^{x^3/3}\,e^{x^4/4}(1+y)} \]