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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
Choose the correct answer from the given four option.The solution of $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\text{e}^{\text{-x}},\text{ y}(0)$ is:
  • A
    $\text{y}=\text{e}^{\text{x}}(\text{x}-1)$
  • ✓
    $\text{y}=\text{x}\text{e}^{-\text{x}}$
  • C
    $\text{y}=\text{x}\text{e}^{\text{x}}+1$
  • D
    $\text{y}=(\text{x}+1)\text{e}^{-\text{x}}$
✓

Answer

Correct option: B.
$\text{y}=\text{x}\text{e}^{-\text{x}}$
We have, $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=\text{e}^{\text{-x}}$
This is a linear differential equation.
On comparing it with $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{P}\text{y}=\text{Q},$we get
$\text{P}=1,\text{Q}=\text{e}^{-\text{x}}$
$\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{{\int\text{dx}}}​​$
So, the general solution is,
$\text{y}.\text{e}^{\text{x}}=\int\text{e}^{-\text{x}}\text{e}^{\text{x}}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\text{e}^{\text{x}}=\int\text{dx}+\text{C}$
$\Rightarrow\text{y}.\text{e}^{\text{x}}=\text{x}+\text{C}$
Given that when $x = 0$ and $y = 0$
$\Rightarrow0 = 0 + \text{C}$
$\Rightarrow\text{C}=0$
$Eq. (i)$ becomes $\text{y}.\text{e}^{\text{x}}=\text{x}$
$\Rightarrow\text{y}=\text{x}\text{e}^{-\text{x}}$

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