Question

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

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

Answer 1

The correct option is C $\frac{d^{2}x}{d{y}^2}-y{\left(\frac{dx}{dy}\right)}^2=0$

Given $\frac{d^{2}y}{{dx}^2}+y\frac{dy}{dx}=0$......(1)

Now

$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$

$\therefore \frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{1}{\frac{dx}{dy}}\right)=\frac{d}{dy}\left(\frac{1}{\frac{dx}{dy}}\right)\frac{dy}{dx}=-\frac{1}{{\left(\frac{dx}{dy}\right)}^2}\frac{d^{2}x}{{dy}^2}.\frac{1}{\frac{dx}{dy}}$

$\Rightarrow \frac{d^{2}y}{{dx}^2}=-\frac{\frac{d^{2}x}{{dy}^2}}{{\left(\frac{dx}{dy}\right)}^3}$

(substitute in (1)), we get

$\frac{-\frac{d^{2}x}{{dy}^2}}{{\left(\frac{dx}{dy}\right)}^3}+y\frac{dy}{dx}=0\Rightarrow y{\left(\frac{dx}{dy}\right)}^2-\frac{d^{2}x}{{dy}^2}=0$