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

Answer: (C)

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