Question

The general solution of the differential equation: \[ x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx. \]

• A. \( y + \sqrt{x^2 + y^2} = c x^2 \)
• B. \( y + \sqrt{x^2 + y^2} = cx \)
• C. \( x + \sqrt{x^2 + y^2} = cy \)
• D. \( x - \sqrt{x^2 + y^2} = c y^2 \)

Hint:

For solving first-order differential equations, look for separable forms or use appropriate substitutions like \( v = \frac{y}{x} \) to simplify the equation before integrating.

Answer 1

The Correct Option is A

Step 1: Expressing the equation in standard form The given differential equation is: \[ x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx. \] Rearrange terms to obtain: \[ x \, dy = y \, dx + \sqrt{x^2 + y^2} \, dx. \] \[ x \, dy = (y + \sqrt{x^2 + y^2}) \, dx. \] Rewriting in separable form: \[ \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}. \]
Step 2: Using substitution to simplify the equation Define a substitution: \[ v = \frac{y}{x} \Rightarrow y = vx, \quad \frac{dy}{dx} = v + x \frac{dv}{dx}. \] Substituting into the equation: \[ v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x}. \] \[ v + x \frac{dv}{dx} = v + \frac{\sqrt{x^2(1+v^2)}}{x}. \] Simplifying: \[ v + x \frac{dv}{dx} = v + \sqrt{1+v^2}. \] \[ x \frac{dv}{dx} = \sqrt{1+v^2}. \]
Step 3: Solving the integral Rearranging: \[ \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}. \] Integrating both sides: \[ \log |v + \sqrt{1+v^2}| = \log |x| + C. \] Substituting back \( v = \frac{y}{x} \): \[ \log \left| \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} \right| = \log |x| + C. \] \[ \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx. \] Multiplying both sides by \( x \): \[ y + \sqrt{x^2 + y^2} = Cx^2. \] % Final Answer Thus, the correct answer is option (1): \( y + \sqrt{x^2 + y^2} = c x^2 \).