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if y=e^{nx},then\; \left ( \frac{d^{2}y}{dx^{2}} \right )\left ( \frac{d^{2}x}{dy^{2}} \right )  is equal to :

  • Option 1)

    n\, e^{nx}\; \;

  • Option 2)

    n\, e^{-nx}\; \;

  • Option 3)

    1

  • Option 4)

    -n\, e^{-nx}

 

Answers (1)

best_answer

As we learnt in 

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable 
\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )

- wherein

eg:

  \frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0

y=e^{nx}

Then \left(\frac{d^{2}y}{dx^{2}} \right ) \left(\frac{d^{2}x}{dy^{2}} \right )

 \\ \frac{dy}{dx}=n.e^{nx}                                                              Now, \log y=nx        

\frac{d^{2}y}{dx^{2}}=n^{2}.e^{nx}                    .............(i)                            \frac{1}{y}=n.\frac{dx}{dy}

                                                                                        -\frac{1}{y^{2}}=n.\frac{d^{2}x}{dy^{2}}                        ...........(ii)

So that,  \frac{d^{2}y}{dx^{2}}\times \frac{d^{2}x}{dy^{2}}= n^{2}.e^{nx}\times (-\frac{1}{ny^{2}})

= \frac{-n.e^{nx}}{(e^{nx})^{2}}

= -n.e^{-nx}


Option 1)

n\, e^{nx}\; \;

This option is incorrect.

Option 2)

n\, e^{-nx}\; \;

This option is incorrect.

Option 3)

1

This option is incorrect.

Option 4)

-n\, e^{-nx}

This option is correct.

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