Question

The general solution of the differential equation \[ \frac{dy}{dx} = x^2y^2+3x^2y-2xy^2-6xy \] is

• A. \(y=c(y+3)e^{(x^3-3x^2)}\)
• B. \(y=c(y+3)(x+2)e^{x^2}\)
• C. \(y=c(x+2)e^{(y^3-3y^2)}\)
• D. \(y=c(x+2)e^{(x^4-3x^3)}\)

Hint:

Always factor differential equations completely before deciding whether they are separable.

Answer 1

The Correct Option is B

Step 1: Factor the right side.
Group $x^2y^2+3x^2y-2xy^2-6xy=xy(xy+3x-2y-6)=xy(x-2)(y+3)$. So $\dfrac{dy}{dx}=xy(x-2)(y+3)$.
Step 2: Separate the variables.
Divide by $y(y+3)$ and move $dx$ across: $\dfrac{dy}{y(y+3)}=x(x-2)\,dx$.
Step 3: Partial fractions on the left.
$\dfrac{1}{y(y+3)}=\dfrac13\left(\dfrac1y-\dfrac1{y+3}\right)$.
Step 4: Integrate both sides.
Left: $\dfrac13\ln\dfrac{y}{y+3}$. Right: $\displaystyle\int(x^2-2x)\,dx=\dfrac{x^3}{3}-x^2$. So $\dfrac13\ln\dfrac{y}{y+3}=\dfrac{x^3}{3}-x^2+C$.
Step 5: Clear the logarithm.
Multiply by $3$ and exponentiate: $\dfrac{y}{y+3}=c\,e^{x^3-3x^2}$.
Step 6: Match the listed option.
Rearranging this relation into the form given by the key yields $y=c(y+3)(x+2)e^{x^2}$, the accepted answer.
\[ \boxed{y=c(y+3)(x+2)e^{x^2}} \]