Question

If \(y=f(x)\) passing through \((1,2)\) satisfies the differential equation \[ y(1+xy)\,dx - x\,dy = 0, \] then \(f(x)=\)

• A. \(\dfrac{2x}{2-x^2}\)
• B. \(\dfrac{x+1}{x^2+1}\)
• C. \(\dfrac{x-1}{4-x^2}\)
• D. \(\dfrac{4x}{1-2x^2}\)

Hint:

If a differential equation can be written as \[ \frac{dy}{dx}=F\!\left(\frac{y}{x}\right), \] use the substitution \(y=vx\). Always apply the given point to find the constant of integration.

Answer 1

The Correct Option is A

Step 1: Rewrite the given differential equation. \[ y(1+xy)\,dx - x\,dy = 0 \] Rearranging, \[ x\,dy = y(1+xy)\,dx \] \[ \Rightarrow \frac{dy}{dx} = \frac{y(1+xy)}{x} \]
Step 2: Check if the equation is homogeneous. \[ \frac{dy}{dx} = \frac{y}{x} + y^2 \] Let \[ y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Step 3: Substitute \(y=vx\) into the equation. \[ v + x\frac{dv}{dx} = v + v^2x^2 \] \[ x\frac{dv}{dx} = v^2x^2 \] \[ \frac{dv}{v^2} = x\,dx \]
Step 4: Integrate both sides. \[ \int v^{-2}\,dv = \int x\,dx \] \[ -\frac{1}{v} = \frac{x^2}{2} + C \]
Step 5: Substitute back \(v=\dfrac{y}{x}\). \[ -\frac{x}{y} = \frac{x^2}{2} + C \] \[ \Rightarrow \frac{x}{y} = -\frac{x^2}{2} + C' \]
Step 6: Use the given point \((1,2)\). \[ \frac{1}{2} = -\frac{1}{2} + C' \Rightarrow C' = 1 \]
Step 7: Write the final equation. \[ \frac{x}{y} = 1 - \frac{x^2}{2} \] \[ y = \frac{x}{1-\frac{x^2}{2}} = \frac{2x}{2-x^2} \]
Hence, \[ \boxed{f(x)=\dfrac{2x}{2-x^2}} \]