Question

The general solution of the differential equation \[ (x + y)y \,dx + (y - x)x \,dy = 0 \] is: 

• A. \( x + y \log(cy) = 0 \)
• B. \( \frac{y}{x} = \log(xy) + c \)
• C. \( x + y \log(cxy) = 0 \)
• D. \( \frac{y}{x} = \log(cxy) \)

Hint:

When solving first-order differential equations, check whether the equation can be rewritten in a separable form. Then, integrate both sides accordingly.

Answer 1

The Correct Option is C

Step 1: Rewrite the given equation
The given differential equation is: \[ (x + y)y \,dx + (y - x)x \,dy = 0. \] Rearrange it as: \[ \frac{dy}{dx} = \frac{(x+y)y}{(x-y)x}. \] Step 2: Use variable separable method
Rewriting, \[ \frac{(x-y)x}{(x+y)y} dy = dx. \] Separate the variables: \[ \frac{(x-y)}{(x+y)} dy = \frac{y}{x} dx. \] Integrating both sides: \[ \int \frac{(x-y)}{(x+y)} dy = \int \frac{y}{x} dx. \] Step 3: Solve the integral
Solving, \[ x + y \log(cxy) = 0. \] Step 4: Final Answer
Thus, the general solution of the given differential equation is: \[ \boxed{x + y \log(cxy) = 0}. \]

Answer 2

Step 1: Start with the given differential equation 

The equation is: \[ (x + y)y\,dx + (y - x)x\,dy = 0 \] Rearranged into the form \( \frac{dy}{dx} = \dots \): \[ (x + y)y + (y - x)x \cdot \frac{dy}{dx} = 0 \Rightarrow (y - x)x \cdot \frac{dy}{dx} = - (x + y)y \Rightarrow \frac{dy}{dx} = \frac{-(x + y)y}{(y - x)x} \] Multiply numerator and denominator by \( -1 \): \[ \frac{dy}{dx} = \frac{(x + y)y}{(x - y)x} \]

Step 2: Use substitution to simplify

Let’s try the substitution: \[ u = xy \Rightarrow \text{Then } \frac{du}{dx} = y + x\frac{dy}{dx} \] From earlier: \[ \frac{dy}{dx} = \frac{(x + y)y}{(x - y)x} \Rightarrow x\frac{dy}{dx} = \frac{(x + y)y}{x - y} \Rightarrow \frac{du}{dx} = y + \frac{(x + y)y}{x - y} \] Too complex — try **homogeneous substitution** instead: Let: \[ y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \] Substitute into the equation: \[ \frac{dy}{dx} = \frac{(x + vx)(vx)}{(x - vx)x} = \frac{x(1 + v) \cdot vx}{x(1 - v)} = \frac{v(1 + v)}{1 - v} \] From chain rule: \[ v + x\frac{dv}{dx} = \frac{v(1 + v)}{1 - v} \Rightarrow x\frac{dv}{dx} = \frac{v(1 + v)}{1 - v} - v \Rightarrow x\frac{dv}{dx} = \frac{v(1 + v - (1 - v))}{1 - v} = \frac{v(2v)}{1 - v} = \frac{2v^2}{1 - v} \] So: \[ \frac{1 - v}{2v^2} dv = \frac{dx}{x} \] Now integrate both sides.

Step 3: Integrate both sides

Left side: \[ \int \left( \frac{1}{2v^2} - \frac{1}{2v} \right) dv = \int \left( \frac{1}{2v^2} - \frac{1}{2v} \right) dv = -\frac{1}{2v} - \frac{1}{2} \log |v| \] Right side: \[ \int \frac{dx}{x} = \log |x| \] So: \[ -\frac{1}{2v} - \frac{1}{2} \log |v| = \log |x| + \log |C| \Rightarrow -\frac{1}{2v} - \frac{1}{2} \log |v| = \log |Cx| \] Multiply both sides by 2: \[ -\frac{1}{v} - \log |v| = 2 \log |Cx| \] Recall that \( v = \frac{y}{x} \Rightarrow \frac{1}{v} = \frac{x}{y} \) So: \[ -\frac{x}{y} - \log \left|\frac{y}{x} \right| = 2 \log |Cx| \] Rearranged and exponentiated leads to an expression of the form: \[ x + y \log (cxy) = 0 \]

Final Answer:

\[ \boxed{x + y \log(cxy) = 0} \]