Question

If $y$ is a function of $x$ then $\frac{d^2 y}{d x^2}+y \frac{d y}{d x}=0$. If $x$ is a function of $y$ then the equation becomes -

• A. $\frac{d^2 x}{d y^2}+x \frac{d x}{d y}=0$
• B. $\frac{d^2 x}{d y^2}+y\left(\frac{d x}{d y}\right)^3=0$
• C. $\frac{ d ^2 x }{ dy ^2}- y \left(\frac{ dx }{ dy }\right)^2=0$
• D. $\frac{d^2 x}{d y^2}-x\left(\frac{d x}{d y}\right)^2=0$

Answer 1

Answer: (C)

$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}} $
$\Rightarrow \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{1}{d x / d y}\right)=\frac{d}{d y}\left(\frac{1}{d x / d y}\right) \frac{d y}{d x} $
$=-\frac{1}{(d x / d y)^2} \cdot \frac{d^2 x}{d y^2} \cdot \frac{d y}{d x}$
Now put the value of $\frac{ dy }{ dx } \& \frac{ d ^2 y }{ dx ^2}$ in
$\frac{d^2 y}{d x^2}+y \frac{d y}{d x}=0$
On solving we get $\frac{d^2 x}{d y^2}-y\left(\frac{d x}{d y}\right)^2=0$