Question
The general solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}+\text{y}}\text{is}$
  1. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
  2. $\text{e}^{\text{x}}+\text{e}^{\text{y}}=\text{C}$
  3. $\text{e}^{-\text{x}}+\text{e}^{\text{y}}=\text{C}$
  4. $\text{e}^{-\text{x}}+\text{e}^{-\text{y}}=\text{C}$

Answer

  1. $\text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C}$
The given differential equation is
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x+y}}\ \ \text{or}\ \ \frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}.\text{e}^{\text{y}}$
Separating the variables, we get,
$\frac{1}{\text{e}^{\text{y}}}\text{dy}=\text{e}^\text{x}\ \text{dx}$
$\text{Integrating},\ \int\text{e}^{-\text{y}}\text{dy}=\int\text{e}^\text{x}\ \text{dx}$
$\therefore\ \frac{\text{e}^{-\text{y}}}{-1}=\text{e}^\text{x}+\text{c}'$
$\therefore\ -\text{e}^{-\text{y}}=\text{e}^\text{x}+\text{c}'\ \ \text{or}\ \ \text{e}^\text{x}+\text{e}^{-\text{y}}=-\text{c}'$
$\therefore\ \text{e}^{\text{x}}+\text{e}^{-\text{y}}=\text{C},$ which is required solution.

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