If e^{y}+xy=e, the ordered pair \left ( \frac{\mathrm{d} y}{\mathrm{d} x},\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}} \right ) at x=0 is equal to : 

  • Option 1)

    \left ( -\frac{1}{e},-\frac{1}{e^{2}} \right ) 

  • Option 2)

      \left ( \frac{1}{e},-\frac{1}{e^{2}} \right ) 

  • Option 3)

        \left (- \frac{1}{e},\frac{1}{e^{2}} \right )  

  • Option 4)

    \left ( \frac{1}{e},\frac{1}{e^{2}} \right )

Answers (1)

e^y+xy=e

e^{y}\frac{\mathrm{d} y}{\mathrm{d} x}+x\frac{\mathrm{d} y}{\mathrm{d} x}+y=0

\frac{\mathrm{d} y}{\mathrm{d} x}\left ( e^{y} +x \right )=-y\: \: \: \: \: ,\: \: \: \: \frac{\mathrm{d} y}{\mathrm{d} x} |_{\left ( 0,1 \right )}=\frac{-1}{e^{1}+0}=\frac{-1}{e}

\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-y}{e^{y}+x}

Again differentiate w.r.t.   x

e^{y}\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\frac{\mathrm{d} y}{\mathrm{d} x}\cdot e^{y}\frac{\mathrm{d} y}{\mathrm{d} x}+x\cdot \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{\mathrm{d} y}{\mathrm{d} x}=0

\left ( x+e^{y} \right )\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( \frac{\mathrm{d} y}{\mathrm{d} x}\right )^{2}e^{y}+2\frac{\mathrm{d} y}{\mathrm{d} x}=0

\left ( 0+e^{1} \right )\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\left ( -\frac{1}{e} \right )^{2}e^{1}+2\cdot \frac{-1}{e}=0

e\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}+\frac{1}{e}-\frac{2}{e}=0

\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=\frac{1}{e^{2}}

\left ( \frac{\mathrm{d} y}{\mathrm{d} x},\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \right )=\left ( -\frac{1}{e},\frac{1}{e^{2}} \right )


Option 1)

\left ( -\frac{1}{e},-\frac{1}{e^{2}} \right ) 

Option 2)

  \left ( \frac{1}{e},-\frac{1}{e^{2}} \right ) 

Option 3)

    \left (- \frac{1}{e},\frac{1}{e^{2}} \right )  

Option 4)

\left ( \frac{1}{e},\frac{1}{e^{2}} \right )

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