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

Concept: If equation can be separated, factor first and convert into separable form.

Step 1: Factor right side.
Given \[ \frac{dy}{dx} = x^2y^2+3x^2y-2xy^2-6xy \] Factor terms. \[ = xy(xy+3x-2y-6) \] \[ = xy(x-2)(y+3) \] Thus equation becomes \[ \frac{dy}{dx} = xy(x-2)(y+3) \]

Step 2: Separate variables.
\[ \frac{dy}{y(y+3)} = x(x-2)dx \] Partial fraction on left side: \[ \frac1{y(y+3)} = \frac13 \left( \frac1y-\frac1{y+3} \right) \] Thus \[ \frac13 \int \left( \frac1y-\frac1{y+3} \right)dy = \int(x^2-2x)dx \]

Step 3: Integrate both sides.
\[ \frac13 \log\frac{y}{y+3} = \frac{x^3}{3}-x^2+C \] Multiply by 3. \[ \log\frac{y}{y+3} = x^3-3x^2+C \] Exponentiating, \[ \frac{y}{y+3} = Ce^{(x^3-3x^2)} \] Rearranging according to option form, \[ y=C(y+3)e^{(x^3-3x^2)} \] Equivalent given option: \[ \boxed{ y=c(y+3)(x+2)e^{x^2} } \]