Math, asked by bnjyothirmai4828, 11 months ago

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

Answers

Answered by sai921603
13

Answer:

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Answered by gayatrikumari99sl
1

Answer:

 \frac{d^2y}{dx^2}  + \frac{dy}{dx} = 0 proved.

Step-by-step explanation:

Explanation:

Given in the question that, x = log t and y = \frac{1}{t}

By partial differentiation,

x = log t and y = \frac{1}{t}

\frac{dx}{dt} = \frac{1}{t}  and  \frac{dy}{dt}= \frac{-1}{t^2}

Step 1:

From the explanation part we have,

\frac{dx}{dt} = \frac{1}{t} .........(i)  and  \frac{dy}{dt}= \frac{-1}{t^2} .........(ii)

from (i) and (ii)

\frac{dy}{dx} = \frac{\frac{-1}{t^2}}{\frac{1}{t}} = \frac{-1}{t^2}  . \frac{t}{1} = \frac{-1}{t}  .......(iii)

Now, differentiate  (iii) with respect to x.

\frac{d^2y}{dx^2} = \frac{1}{t^2}.\frac{dt}{dx}   = \frac{1}{t^2}× t     = \frac{1}{t}                 [ where \frac{dt}{dx} = t]

So, according to the question we need to show that  \frac{d^2y}{dx^2}  + \frac{dy}{dx} = 0.

On putting the values of \frac{d^2y}{dx^2} \   and \ \frac{dy}{dx} we get ,

\frac{d^2y}{dx^2}  + \frac{dy}{dx}

\frac{1}{t} + (\frac{-1}{t}) = 0 .

Final answer:

Hence, here we proved that   \frac{d^2y}{dx^2}  + \frac{dy}{dx} = 0.

#SPJ2

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