Question

If x=logt and y=1/t then prove that d2y/dx2 + dy/dx=0

Answer 1

$\textbf{Given:}$

$x=\logt\;\text{and}\;y=\frac{1}{t}$

$\textbf{To prove:}$

$\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0$

$x=\log\,t\;\implies\;\dfrac{dx}{dt}=\frac{1}{t}$

$y=t^{-1}\;\implies\;\dfrac{dy}{dt}=-t^{-2}$

$\text{Now,}$

$\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\dfrac{dy}{dx}=\dfrac{-t^{-2}}{\frac{1}{t}}$

$\dfrac{dy}{dx}=-t^{-2}{\times}t$

$\implies\bf\dfrac{dy}{dx}=-t^{-1}$

$\text{Differentiate with respect to x}$

$\dfrac{d^2y}{dx^2}=\dfrac{d(-t^{-1})}{dx}$

$\dfrac{d^2y}{dx^2}=\dfrac{d(-t^{-1})}{dt}\,\dfrac{dt}{dx}$

$\dfrac{d^2y}{dx^2}=(t^{-2})\,\dfrac{1}{\frac{dx}{dt}}$

$\dfrac{d^2y}{dx^2}=(t^{-2})\,\dfrac{1}{\frac{1}{t}}$

$\dfrac{d^2y}{dx^2}=(t^{-2})t$

$\dfrac{d^2y}{dx^2}=t^{-1}$

$\text{Consider}$

$\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}$

$=t^{-1}+(-t^{-1})$

$=t^{-1}-t^{-1}$

$=\frac{1}{t}-\frac{1}{t}$

$=0$

$\implies\bf\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0$

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