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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The solution of the differention $\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$ is :
  • A
    $(\text{x}+\text{y})\text{e}^{\text{x}+\text{y}}=0$
  • B
    $(\text{x}+\text{C})\text{e}^{\text{x}+\text{y}}=0$
  • C
    $(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}=1$
  • ✓
    $(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$
✓

Answer

Correct option: D.
$(\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$
We have,
$\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x}+\text{y}}$
Let $\text{x}+\text{y}=\text{u}$
$\Rightarrow 1+\frac{\text{dy}}{\text{dx}}=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}+1=\frac{\text{du}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{u}}$
$\Rightarrow \text{e}^{-\text{u}}\text{du}=\text{dx}$
Intergrating both sides, we get
$\Rightarrow \text{e}^{-\text{u}}=\text{x}-\text{C}$
$\Rightarrow -1=\text{e}^{-\text{u}}(\text{x}-\text{C})$
$\Rightarrow (\text{x}-\text{C})\text{e}^{\text{x}+\text{y}}+1=0$

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