MCQ
The solution of the differential equation $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}+\text{x}\ \tan\frac{\text{y}}{\text{x}}$ is:
  • $\sin\frac{\text{x}}{\text{y}}=\text{x}+\text{C}$
  • B
    $\sin\frac{\text{y}}{\text{x}}=\text{Cx}$
  • C
    $\sin\frac{\text{x}}{\text{y}}=\text{Cy}$
  • D
    $\sin\frac{\text{y}}{\text{x}}=\text{Cy}$

Answer

Correct option: A.
$\sin\frac{\text{x}}{\text{y}}=\text{x}+\text{C}$
We have,
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}+\text{x}\ \tan\frac{\text{y}}{\text{x}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\tan\frac{\text{y}}{\text{x}}\ ...(\text{i})$
Let $\text{y}=\upsilon\text{x}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}$
Putting both value in (i)
$\upsilon+\text{x}\frac{\text{d}\upsilon}{\text{dx}}=\upsilon+\tan\upsilon$
$\Rightarrow \frac{\text{d}\upsilon}{\tan\upsilon}=\frac{\text{dx}}{\text{x}}$
Integrating both sides, we get
$\log\sin\upsilon=\log\text{x}+\log\text{C}$
$\Rightarrow \log\frac{\sin\upsilon}{\text{x}}=\log\text{C}$
$\Rightarrow \frac{\sin\upsilon}{\text{x}}=\text{C}$
$\Rightarrow\sin\upsilon=\text{Cx}$
$\Rightarrow\sin(\frac{\text{y}}{\text{x}})=\text{Cx}$

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