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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The general solution of the differential equation $\text{e}^\text{x}\ \text{dy}+(\text{y e}^\text{x}+2\text{x})\text{dx}=0\ \text{is}$
  • A
    $\text{x e}^\text{y}+\text{x}^2=\text{C}$
  • B
    $\text{x e}^\text{y}+\text{y}^2=\text{C}$
  • ✓
    $\text{y e}^\text{x}+\text{x}^{2}=\text{C}$
  • D
    $\text{y e}^\text{y}+\text{x}^2=\text{C}$
✓

Answer

Correct option: C.
$\text{y e}^\text{x}+\text{x}^{2}=\text{C}$
The given differential equation is
$\text{e}^\text{x}\ \text{dy}+(\text{y e}^\text{x}+2\text{x})\text{dx}=0$
or$\ \ \text{e}^\text{x}\frac{\text{dy}}{\text{dx}}+\text{y e}^\text{x}+2\text{x}=0\ \ $or$\ \ \frac{\text{dy}}{\text{dx}}+\text{y}+\frac{2\text{x}}{\text{e}^\text{x}}=0$ or $\ \ \frac{\text{dy}}{\text{dx}}+\text{y}=-2\text{x e}^{-\text{x}}$
Comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$
we get$,\ \text{P}=1,\ \text{Q}=-2\text{x e}^{-\text{x}}$
$\therefore\ \ \int\text{P dx}=\int1\ \text{dx}=\text{'x},\ \text{e}^{\int\text{P dx}}=\text{e}^\text{x}$
Solution of given differential equation is
$\text{y e}^{\int\text{P dx}}=\int\text{Q e}^{\int\text{P dx}}+\text{C}$ $\text{or}\ \ \text{y e}^\text{x}=\int(-2\text{x e}^\text{x}).\text{e}^{-\text{x}}\ \text{dx}+\text{C}$
$\text{or}\ \ \text{y e}^\text{x}=-2\int\text{x dx}+\text{C}\ \ \text{or}\ \ \text{y e}^\text{x}=-\text{x}^2+\text{C}$
$\text{or}\ \ \text{y e}^\text{x}+\text{x}^2=\text{C}$
$\therefore (C)$ is correct answer.

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