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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 options.The solution of $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}},$ y(0) = 0 is:
  • A
    $\text{y}=\text{e}^{-\text{x}}(\text{x}-1)$
  • B
    $\text{y}=\text{x}\text{e}^{\text{x}}$
  • C
    $\text{y}=\text{x}\text{e}^{-\text{x}}+1$
  • ✓
    $\text{y}=\text{x}\text{e}^{-\text{x}}$
✓

Answer

Correct option: D.
$\text{y}=\text{x}\text{e}^{-\text{x}}$
We have, $\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{-\text{x}}$
Which is a linear differential equation.

Here, $\text{P}=1$ and $\text{Q}=\text{e}^{-\text{x}}$

$\text{I.F.}=\text{e}^{\int\text{dx}}=\text{e}^{\text{x}}$

The general solution is

$\text{y}.\text{e}^{-\text{x}}=\int\text{e}^{-\text{x}}.\text{e}^{\text{x}}\ \text{dx}+\text{C}$

$\Rightarrow\text{y}\text{e}^{\text{x}}=\int\text{dx}+\text{C}$

$\Rightarrow\text{y}\text{e}^{-\text{x}}=\text{x}+\text{C}\ ....(\text{i})$

Given, when x = 0 and y = 0

$\Rightarrow0=0+\text{C}$

$\Rightarrow\text{C}=0$

Eq, (i) resuces to $\text{y}.\text{e}^{-\text{x}}=\text{x}$ or $\text{y}=\text{x}\text{e}^{-\text{x}}.$

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