Question

Find the general solution of the differential equation: $$ (xy + y^2)dx - (x^2 - 2xy)dy = 0. $$ is 

• A. \( c x y^2 = e^{x/y} \)
• B. \( c x y^2 e^{x/y} = 1 \)
• C. \( c x y e^{x/y} = 1 \)
• D. \( c x y = e^{x/y} \)

Hint:

To solve homogeneous differential equations, use the substitution \( u = \frac{x}{y} \), express \( x \) in terms of \( y \), and differentiate accordingly to separate variables before integration.

Answer 1

The Correct Option is B

Step 1: Identifying the Type of Differential Equation
The given differential equation is: \[ (xy + y^2)dx - (x^2 - 2xy)dy = 0. \] Rewriting in the form: \[ \frac{dx}{dy} = \frac{x^2 - 2xy}{xy + y^2}. \] Step 2: Separating the Variables Rearranging: \[ \frac{dx}{dy} = \frac{x^2 - 2xy}{xy + y^2}. \] Dividing numerator and denominator by \( y^2 \): \[ \frac{dx}{dy} = \frac{\frac{x^2}{y^2} - 2\frac{x}{y}}{\frac{x}{y} + 1}. \] Let \( u = \frac{x}{y} \), so \( x = uy \) and differentiating: \[ \frac{dx}{dy} = u + y \frac{du}{dy}. \] Step 3: Solving for \( u \)
Substituting in terms of \( u \), solving, and integrating both sides, we get: \[ c \cdot x y^2 e^{x/y} = 1. \] Step 4: Conclusion
Thus, the general solution of the given differential equation is: \[ \mathbf{c \cdot x y^2 e^{x/y} = 1}. \]