Solve the following differential equation: dx + xdy=e^ -y ^2y dy

Question

Solve the following differential equation:
$\text{dx + xdy}=\text{e}^{-\text{y}}\sec^2\text{y dy}$

✓

Answer

Here, $\text{dx + xdy}=\text{e}^{-\text{y}}\sec^2\text{y dy}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}+\text{x}=\text{e}^{-\text{y}}\sec^2\text{y}$
It is a linear differential equation. Comparing it with,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=1,\text{Q}=\text{e}^{-\text{y}}\sec^2\text{y}$
I.F. $=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\text{dy}}$
$=\text{e}^{\text{y}}$
Solution of the question is given by,
$\text{x}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dy + C}$
$\text{xe}^{\text{y}}=\int\text{e}^{-\text{y}}\sec^2\text{ye}^{\text{y}}\text{dy + C}$
$=\int\sec^2\text{ydy}+\text{C}$
$\text{xe}^{\text{y}}=\int\tan\text{y + C}$
$\text{x}=\text{e}^{-{\text{y}}}(\tan\text{y + C})$

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