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: Use the Structure
Let $w=xe^{-y/x}$. Since $w=a+bx$ is linear in $x$, $\dfrac{d^{2}w}{dx^{2}}=0$.

Step 2: Compute w Prime
$w'=e^{-y/x}\dfrac{x-p}{x}$ with $p=xy'-y$. Differentiating once more, the exponential factor gives $-\dfrac{p}{x^{2}}$ times $w'$, and $\dfrac{x-p}{x}$ has derivative $\dfrac{p}{x^{2}}-y''$.

Step 3: Set to Zero
$0=-\dfrac{p(x-p)}{x^{3}}+\dfrac{p}{x^{2}}-y''\Rightarrow y''=\dfrac{p^{2}}{x^{3}}$, which is option (B).

Final Answer:
Eliminating a and b gives option (B). \[ \boxed{\text{(B) }x^{3}\dfrac{d^{2}y}{dx^{2}}=\left(x\dfrac{dy}{dx}-y\right)^{2}} \]