Question

The solution of the differential equation \( y^2 dx + (x^2 - xy + y^2) dy = 0 \) is

• A. \( \tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0 \)
• B. \( 2 \tan^{-1}\left(\frac{x}{y}\right) + \ln x + C = 0 \)
• C. \( \ln(v + \sqrt{x^2 + y^2}) + \ln y + C = 0 \)
• D. \( \ln(x + \sqrt{x^2 + y^2}) + C = 0 \)

Hint:

When solving a differential equation using substitution, make sure to fully simplify and separate the variables before performing integration.

Answer 1

The Correct Option is A

From the question we have given the differential equation is: \[ y^2 \, dx + (x^2 - xy + y^2) \, dy = 0. \] Step 1: Divide through by \( y^2 \). \[ \frac{dx}{dy} + \frac{x^2 - x}{y^2} + 1 = 0. \] Step 2: Use the substitution \( x = vy \), where \( v = \frac{x}{y} \). Then, \[ \frac{dx}{dy} = v + y \frac{dv}{dy}. \] Substitute this into the equation: \[ v + y \frac{dv}{dy} + v^2 - v + 1 = 0. \] Simplify the equation: \[ y \frac{dv}{dy} = -(1 + v^2). \] Step 3: Separate variables and integrate. \[ \frac{dv}{1 + v^2} = -\frac{dy}{y}. \] Integrating both sides: \[ \tan^{-1} v = -\ln y + C, \] where \( C \) is the constant of integration. Step 4: Substitute \( v = \frac{x}{y} \) back into the equation. \[ \tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0. \] Final Answer: \[ \boxed{\tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0.} \]