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

The given differential equation is:\[y^2 \, dx + (x^2 - xy + y^2) \, dy = 0.\]Step 1: Divide the equation by \( y^2 \).\[\frac{dx}{dy} + \frac{x^2 - x}{y^2} + 1 = 0.\]Step 2: Apply the substitution \( x = vy \), where \( v = \frac{x}{y} \). This implies\[\frac{dx}{dy} = v + y \frac{dv}{dy}.\]Substituting this into the equation yields:\[v + y \frac{dv}{dy} + v^2 - v + 1 = 0.\]Simplifying this equation gives:\[y \frac{dv}{dy} = -(1 + v^2).\]Step 3: Separate the variables and integrate.\[\frac{dv}{1 + v^2} = -\frac{dy}{y}.\]Integrating both sides leads to:\[\tan^{-1} v = -\ln y + C,\]where \( C \) is the constant of integration.Step 4: Substitute \( v = \frac{x}{y} \) back into the integrated equation.\[\tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0.\] Final Answer:\[\boxed{\tan^{-1}\left(\frac{x}{y}\right) + \ln y + C = 0.}\]