Question

The differential equation which represents the family of curves \((a+bx)e^{y/x}=x\), (where \(a\) and \(b\) are arbitrary constants); is

• A. \(x\frac{d^{2}y}{dx^{2}}=x\frac{dy}{dx}-y\)
• B. \(x^{3}\frac{d^{2}y}{dx^{2}}=\left(x\frac{dy}{dx}-y\right)^{2}\)
• C. \(x^{2}\frac{d^{2}y}{dx^{2}}=x\frac{dy}{dx}-y\)
• D. \(x^{3}\frac{d^{2}y}{dx^{2}}=\left(x\frac{dy}{dx}-y\right)^{3}\)

Hint:

Rewrite as a + bx = x e^(-y/x) and differentiate twice to eliminate a and b.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
There are two arbitrary constants, so differentiate twice. Rewrite as \(a+bx=xe^{-y/x}\).

Step 2: First Derivative
Differentiating, \(b=e^{-y/x}\left[1-\dfrac{xy'-y}{x}\right]=e^{-y/x}\,\dfrac{x+y-xy'}{x}\).

Step 3: Second Derivative
The derivative of \(b\) is zero. Writing \(p=xy'-y\), this gives \(-\dfrac{p(x-p)}{x^{3}}+\dfrac{p}{x^{2}}-y''=0\), because \(x+y-xy'=x-p\).

Step 4: Simplify
The terms \(\dfrac{p}{x^{2}}\) cancel, leaving \(y''=\dfrac{p^{2}}{x^{3}}\), i.e. \(x^{3}\dfrac{d^{2}y}{dx^{2}}=\left(x\dfrac{dy}{dx}-y\right)^{2}\), option (B).

Final Answer:
The differential equation is x cubed y double prime equals (x y prime minus y) squared. \[ \boxed{\text{(B) }x^{3}\dfrac{d^{2}y}{dx^{2}}=\left(x\dfrac{dy}{dx}-y\right)^{2}} \]