Question

The solution of the differential equation \[ x dy - y dx = \sqrt{x^2 + y^2} dx \] when \( y(\sqrt{3}) = 1 \) is:

• A. \( y^2 + \sqrt{x^2 + y^2} = x^2 \)
• B. \( 5y - \sqrt{x^2 + y^2} = x^2 \)
• C. \( y + \sqrt{x^2 + y^2} = x^2 \)
• D. \( 5y^2 - \sqrt{x^2 + y^2} = x \)

Hint:

When solving exact equations, check if it can be rewritten in separable form.

Answer 1

The Correct Option is C

Step 1: Rewrite the Given Differential Equation We start with the given equation: \[ x \frac{dy}{dx} - y = \sqrt{x^2 + y^2} \] Rearrange it into standard form: \[ \frac{dy}{dx} = \frac{\sqrt{x^2 + y^2} + y}{x} \] Step 2: Transforming into a Suitable Form Introduce a substitution: \[ y = vx \] where \( v = \frac{y}{x} \), so that \( \frac{dy}{dx} = v + x \frac{dv}{dx} \). Substituting into the differential equation: \[ v + x \frac{dv}{dx} = \frac{\sqrt{x^2 + v^2 x^2} + v x}{x} \] Simplify: \[ v + x \frac{dv}{dx} = \frac{\sqrt{x^2(1+v^2)} + v x}{x} \] \[ v + x \frac{dv}{dx} = \frac{x\sqrt{1+v^2} + v x}{x} \] \[ v + x \frac{dv}{dx} = \sqrt{1+v^2} + v \] Canceling \( v \) from both sides: \[ x \frac{dv}{dx} = \sqrt{1+v^2} \] Step 3: Separating Variables and Integrating Rearrange to separate \( v \) and \( x \): \[ \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x} \] Integrating both sides: \[ \int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x} \] Using the standard integral formula: \[ \ln |v + \sqrt{1+v^2}| = \ln |x| + C \] Step 4: Substituting Back \( v = \frac{y}{x} \) \[ \ln \left| \frac{y}{x} + \sqrt{1+ \frac{y^2}{x^2}} \right| = \ln |x| + C \] \[ \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx \] Multiplying by \( x \) to clear fractions: \[ y + \sqrt{x^2 + y^2} = C x^2 \] Step 5: Finding the Constant \( C \) Given \( y(\sqrt{3}) = 1 \), substitute \( x = \sqrt{3} \) and \( y = 1 \): \[ 1 + \sqrt{3 + 1} = C (3) \] \[ 1 + 2 = 3C \] \[ C = 1 \] Step 6: Final Solution Substituting \( C = 1 \): \[ y + \sqrt{x^2 + y^2} = x^2 \]