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STD 12 - 9. differential equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 9. differential equations4 Marks
MCQ
If $\mathrm{y}=\mathrm{y}(\mathrm{x})$ is the solution of the differential equation, $\mathrm{e}^{\mathrm{y}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}-1\right)=\mathrm{e}^{\mathrm{x}}$ such that $\mathrm{y}(0)=0,$ then $\mathrm{y}(1)$ is equal to 
  • A
    $2+\log _{e} 2$
  • B
    $2e$
  • C
    $\log _{e} 2$
  • ✓
    $1+\log _{e} 2$
✓

Answer

Correct option: D.
$1+\log _{e} 2$
d
$e^{y} \frac{d y}{d x}-e^{y}=e^{x},$ Let $e^{y}=t$

$\Rightarrow \mathrm{e}^{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{d t}{d x}-t=e^{x}$

I.F. $=e^{\int-d x}=e^{-x}$

$\mathrm{t} \mathrm{e}^{-\mathrm{x}}=\mathrm{x}+\mathrm{c} \Rightarrow \mathrm{e}^{\mathrm{y}-\mathrm{x}}=\mathrm{x}+\mathrm{c}$

$y(0)=0 \Rightarrow c=1$

$\mathrm{e}^{\mathrm{y}-\mathrm{x}}=\mathrm{x}+1 \Rightarrow \mathrm{y}(1)=1+\log _{\mathrm{e}} 2$

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