Question

The solution of the differential equation \[ \frac{dy}{dx} = \frac{y + x \tan \left( \frac{y}{x} \right)}{x}. \] \[ \sin\frac{y}{x} = \]

• A. \( cx^2 \)
• B. \( cx \)
• C. \( cx^3 \)
• D. \( cx^4 \)

Hint:

For such differential equations, look for simple solutions that involve the function and its derivatives. Use separation of variables if possible.

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{dy}{dx} = \frac{y + x \tan \left( \frac{y}{x} \right)}{x}. \] Let \( u = \frac{y}{x} \), so \( y = ux \). Now, differentiate \( y = ux \) with respect to \( x \): \[ \frac{dy}{dx} = u + x \frac{du}{dx}. \] Substitute this into the original equation: \[ u + x \frac{du}{dx} = \frac{ux + x \tan(u)}{x}. \] Simplifying: \[ u + x \frac{du}{dx} = u + \tan(u). \] Subtract \( u \) from both sides: \[ x \frac{du}{dx} = \tan(u), \] and divide both sides by \( x \): \[ \frac{du}{dx} = \frac{\tan(u)}{x}. \] Now, separate variables: \[ \frac{du}{\tan(u)} = \frac{dx}{x}. \] Integrating both sides: \[ \int \frac{1}{\tan(u)} du = \int \frac{1}{x} dx. \] The integral of \( \frac{1}{\tan(u)} \) is \( \ln|\sin(u)| \), and the integral of \( \frac{1}{x} \) is \( \ln|x| \): \[ \ln|\sin(u)| = \ln|x| + C. \] Exponentiating both sides: \[ |\sin(u)| = Cx. \] Finally, substitute \( u = \frac{y}{x} \): \[ \sin\left(\frac{y}{x}\right) = Cx. \] Thus, the correct answer is \( cx \).