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Need clarity, kindly explain! - Differential equations - JEE Main

If   y=(x+\sqrt{1+x^{2}})^{n},\; then\; (1+x^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx}     is

  • Option 1)

    n^{2}y\;

  • Option 2)

    \; -n^{2}y\;

  • Option 3)

    \; -y\;

  • Option 4)

    \; 2x^{2}y

 
Answers (1)
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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=(x+\sqrt {1+x^2})^{n}

\frac{dy}{dx}= n(x+\sqrt {1+x^{2}})^{n-1}\:\:\times \left ( 1+ \frac{2x}{2\sqrt{1+x^{2}}} \right )

=\frac{n(x+ \sqrt{1+x^{2})}^{n-1}\:\times\: (x+\sqrt{17 x^{2}})}{\sqrt{1+x^{2}}}

\frac{dy}{dx}\:\:\sqrt{1+x^{2}}= n (x+ \sqrt{1+x^{2}})^{n}=ny

\frac{d^{2}y}{dy^{2}}\:\sqrt{1+x^{2}}+\frac{2x}{2 \sqrt {1+x^{2}}}\:\:\:\:\frac{dy}{dx}=n.\frac{dy}{dx}

\frac{d^{2}y}{dy^{2}}\:(1+x^{2} ) +x\:\frac{dy}{dx}=n \sqrt{1+ x^{2}}\:\:\frac{dy}{dx}

=n \times ny=n^{2}

 


Option 1)

n^{2}y\;

This option is correct.

Option 2)

\; -n^{2}y\;

This option is incorrect.

Option 3)

\; -y\;

This option is incorrect.

Option 4)

\; 2x^{2}y

This option is incorrect.

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