MCQ
જો ${e^y} + xy = e$, તો ${{{d^2}y} \over {d{x^2}}}$ એ $x = 0$ આગળ મેળવો.
- A${1 \over e}$
- ✓${1 \over {{e^2}}}$
- C${1 \over {{e^3}}}$
- Dએકપણ નહીં
Differentiating $w.r.t.\,\, x,$ we get
${e^y}\frac{{dy}}{{dx}} + y + x\frac{{dy}}{{dx}} = 0$ .....$(i)$
Differentiating $w.r.t. \,\,x,$ we get
${e^y}\frac{{{d^2}y}}{{d{x^2}}} + {e^y}{\left( {\frac{{dy}}{{dx}}} \right)^2} + 2\frac{{dy}}{{dx}} + x\frac{{{d^2}y}}{{d{x^2}}} = 0$ .....$(ii)$
Putting $x = 0 $ in ${e^y} + xy = e,$we get $y = 1$
Putting $x = 0,\;\;y = 1$ in (i), we get $e\frac{{dy}}{{dx}} + 1 = 0$
==> $\frac{{dy}}{{dx}} = - \frac{1}{e}$
Putting $x = 0,\;y = 1,\;\frac{{dy}}{{dx}} = - \frac{1}{e}$ in $(ii),$ we get
${e} \frac{{{d^2}y}}{{d{x^2}}} + e \cdot \frac{1}{e^2} - \frac{2}{e} + {0} \cdot \frac{{{d^2}y}}{{d{x^2}}} = 0$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{1}{e^2}$
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