Question

The general solution of the differential equation \[ y(2x+y)\,dx=x(x+y)\,dy \] is

• A. \[ \log\left(\frac{cx^2}{y}\right)+\frac{y}{x}=0 \]
• B. \[ \log\left(\frac{y}{cx^2}\right)+\frac{y}{x}=0 \]
• C. \[ \log(cx^2y^2)+\frac{y}{x}=0 \]
• D. \[ \log(cx^2y)+\frac{y}{x}=0 \]

Hint:

Whenever a differential equation is homogeneous, use \[ y=vx \] (or \(x=vy\)). This converts the equation into a separable differential equation in \(v\) and \(x\).

Answer 1

The Correct Option is B

Concept: The given differential equation is homogeneous since all terms are of degree \(2\). Use the substitution \[ y=vx. \] Then \[ \frac{dy}{dx}=v+x\frac{dv}{dx}. \]

Step 1: Rewrite the differential equation. Given, \[ y(2x+y)\,dx=x(x+y)\,dy. \] Hence, \[ \frac{dy}{dx} = \frac{y(2x+y)}{x(x+y)}. \] Substituting \[ y=vx, \] \[ v+x\frac{dv}{dx} = \frac{vx(2x+vx)}{x(x+vx)}. \] \[ = \frac{v(2+v)}{1+v}. \] Therefore, \[ x\frac{dv}{dx} = \frac{v(2+v)}{1+v}-v. \] \[ = \frac{v}{1+v}. \] Thus, \[ \frac{1+v}{v}\,dv = \frac{dx}{x}. \]

Step 2: Integrate both sides. \[ \int\left(\frac1v+1\right)dv = \int\frac{dx}{x}. \] \[ \log|v|+v = \log|x|+C. \]

Step 3: Substitute \(v=\dfrac{y}{x}\). \[ \log\left|\frac{y}{x}\right| + \frac{y}{x} = \log|x|+C. \] \[ \log\left|\frac{y}{x^2}\right| + \frac{y}{x} = C. \] Writing \[ C=-\log c, \] we obtain \[ \log\left(\frac{y}{cx^2}\right) + \frac{y}{x} = 0. \] Therefore, \[ \boxed{ \log\left(\frac{y}{cx^2}\right) + \frac{y}{x} = 0 } \] \[ \boxed{\text{Answer = (B)}} \]