Solve the following differential equation: (x+ )dy= 2y dx

Question

Solve the following differential equation:
$(\text{x}+\tan\text{y})\text{dy}=\sin2\text{y dx}$

✓

Answer

We have,
$(\text{x}+\tan\text{y})\text{dy}=\sin2\text{y dx}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=\text{x cosec }2\text{y}+\frac{1}2\sec^2\text{y}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}-\text{x cosec }2\text{y}=\frac{1}2\sec^2\text{y}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=-\text{cosec }2\text{y}$
$\text{Q}=\frac{1}2\sec^2\text{y}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{-\int\text{cosec }2\text{y }\text{dy}}$
$=\text{e}^{-\frac{1}2\log|\tan{\text{y}}|}=\frac{1}{\sqrt{\tan\text{y}}}$
Multiplying both sides of (1) by $\frac{1}{\sqrt{\tan\text{y}}},$ we get
$\frac{1}{\sqrt{\tan\text{y}}}\Big(\frac{\text{dx}}{\text{dy}}-\text{x cosec 2y}\Big)=\frac{1}2\frac{1}{\sqrt{\tan\text{y}}}\times\sec^2\text{y}$
$\Rightarrow\ \frac{1}{\sqrt{\tan\text{y}}}\frac{\text{dx}}{\text{dy}}-\text{x cosec 2y}\frac{1}{\sqrt{\tan\text{y}}}=\frac{1}2\frac{1}{\sqrt{\tan\text{y}}}\times\sec^2\text{y}$
Integrating both sides with respect to y, we get
$\frac{1}{\sqrt{\tan\text{y}}}\text{x}=\int\frac{1}2\frac{1}{\sqrt{\tan\text{y}}}\times\sec^2\text{y dy + C}$
$\Rightarrow\ \frac{\text{x}}{\sqrt{\tan\text{y}}}=\text{I + C}\ \dots(2)$
where $\text{I}=\int\frac{1}2\frac{1}{\sqrt{\tan\text{y}}}\times\sec^2\text{y dy}$
Putting $\text{t}=\tan\text{y},$ we get
$\text{dt}=\sec^2\text{y dy}$
$\therefore\ \text{I}=\frac{1}2\int\frac{1}{\sqrt{\text{t}}}\times\text{dt}$
$=\sqrt{\text{t}}$
$=\sqrt{\tan\text{y}}$
Putting the value of I in (2), we get
$\frac{\text{x}}{\sqrt{\tan\text{y}}}=\sqrt{\tan\text{y}}+\text{C}$
$\Rightarrow\ \text{x}=\tan\text{y + C}\sqrt{\tan\text{y}}$
Hence, $\text{x}=\tan\text{y + C}\sqrt{\tan\text{y}}$ is the required solution.