Question

If \[ \cos\frac{y}{x}=A\log x+C \] is the general solution of \[ \left(x\sin\frac{y}{x}\right)dy=\left(y\sin\frac{y}{x}-x\right)dx, \] then \(A=\)

• A. \(2\)
• B. \(1\)
• C. \(-1\)
• D. \(-2\)

Hint:

When a differential equation contains \(\frac{y}{x}\), use the substitution \(y=vx\). This often converts the equation into separable form.

Answer 1

The Correct Option is B

Step 1: Write the differential equation in derivative form.
Given, \[ \left(x\sin\frac{y}{x}\right)dy=\left(y\sin\frac{y}{x}-x\right)dx \] Dividing by \(dx\), \[ x\sin\frac{y}{x}\frac{dy}{dx}=y\sin\frac{y}{x}-x \]

Step 2: Use substitution.
Let \[ v=\frac{y}{x} \] Then, \[ y=vx \] So, \[ \frac{dy}{dx}=v+x\frac{dv}{dx} \] Substitute in the equation: \[ x\sin v\left(v+x\frac{dv}{dx}\right)=xv\sin v-x \] Dividing by \(x\), \[ \sin v\left(v+x\frac{dv}{dx}\right)=v\sin v-1 \] \[ v\sin v+x\sin v\frac{dv}{dx}=v\sin v-1 \] Thus, \[ x\sin v\frac{dv}{dx}=-1 \]

Step 3: Separate the variables.
\[ \sin v\,dv=-\frac{dx}{x} \] Integrating both sides, \[ \int \sin v\,dv=-\int \frac{dx}{x} \] \[ -\cos v=-\log x+C \] Multiplying by \(-1\), \[ \cos v=\log x+C \] Since \[ v=\frac{y}{x}, \] we get \[ \cos\frac{y}{x}=\log x+C \]

Step 4: Compare with the given solution.
Given solution is \[ \cos\frac{y}{x}=A\log x+C \] Comparing, \[ A=1 \]

Step 5: Final conclusion.
Hence, \[ \boxed{1} \]