MCQ
Choose the correct answer from the given four option.The general solution of $\text{e}^{\text{x}}\cos\text{ydx}-\text{e}^\text{x}\sin\text{ydy}=0$ is:
  • $\text{e}^{\text{x}}\cos\text{y}=\text{k}$
  • B
    $\text{e}^{\text{x}}\sin\text{y}=\text{k}$
  • C
    $\text{e}^{\text{x}}=\text{k}\cos\text{y}$
  • D
    $\text{e}^{\text{x}}=\text{k}\sin\text{y}$

Answer

Correct option: A.
$\text{e}^{\text{x}}\cos\text{y}=\text{k}$
Given is, $\text{e}^{\text{x}}\cos\text{ydx}-\text{e}^\text{x}\sin\text{ydy}=0$
$\Rightarrow\text{e}^{\text{x}}\cos\text{ydx}=\text{e}^\text{x}\sin\text{ydy}$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}=\tan\text{ydy}$
$\Rightarrow\text{dx}=\tan\text{ydy}$
On integrating both sides, we get
$\text{x}=\log\sec\text{y}+\text{C}$
$\Rightarrow\text{x}-\text{C}=\log\sec\text{y}$
$\Rightarrow\sec\text{y}=\text{e}^{\text{x}-\text{c}}$
$\Rightarrow\frac{1}{\cos\text{y}}=\frac{\text{e}^\text{x}}{\text{e}^\text{c}}$
$\Rightarrow\text{e}^{\text{x}}\cos\text{y}=\text{k}$
$[$where$,\text{K}=\text{e}^\text{c}]$

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