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  2. Help me solve this - Ordinary Differential Equations - BITSAT
# Engineering

If x=e^{t} Sint,y=l^{t} Cost, t is a parameter  then \frac{d^{2}y}{dx^{2}} at (1, 1) os equal to

  • Option 1)

    \frac{-1}{2}

  • Option 2)

    -\frac{1}{4}

  • Option 3)

    0

  • Option 4)

    \frac{1}{2}

 

Answers (1)
D divya.saini

As 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

 

 x=e^{t}\sin t\ \Rightarrow \frac{dx}{dt} =e^{t}\sin t+e^{t}\cos t

y= e^{t}\cos t\ \Rightarrow \frac{dy}{dt}= e^{t}\cos t- e^{t}\sin t

\therefore \frac{dy}{dx} = \frac{\cos t-\sin t}{\cos t+\sin t} =\frac{1-\tan t}{1+\tan t}

                                             =\frac{1-\frac{x}{y}}{1+\frac{x}{y}}

                                             =\frac{y-x}{y+x}\ \ \ \left [ \because \frac{x}{y}= \tan t \right ]

\frac{d^{2}y}{dx^{2}}= \frac{\left ( y+x \right )\left ( \frac{dy}{dv} -1\right )-\left ( y-x \right )\left ( \frac{dy}{dx}+1 \right )}{\left ( y+x \right )^{2}}

         =\frac{\left ( y+x \right )\left ( \frac{y-x-y-x}{y+x} \right ) -\left ( y-x \right )\left ( \frac{y-x+y+x}{y+x} \right ) }{\left ( y+x \right )^{2}}

        \Rightarrow \frac{-2x-2y\left ( \frac{y-x}{y+x} \right )}{\left ( y+x \right )^{2}}

Put

x=1, \ y=1

\therefore \frac{-2\times 1-0}{\left ( 1+1 \right )^{2}}=\frac{-2}{4} =-\frac{1}{2}


Option 1)

\frac{-1}{2}

This option is correct

Option 2)

-\frac{1}{4}

This option is incorrect

Option 3)

0

This option is incorrect

Option 4)

\frac{1}{2}

This option is incorrect

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