For the differential equation in given question find the general solution.

$e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$

Open in App

Solution

Given, $e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$
On separating the varibales , we get $e^{x} t a n y d x = - (1 - e^{x}) s e c^{2} y d y$
$\Rightarrow \frac{e^{x}}{(e^{x} - 1)} d x = \int \frac{s e c^{2} y}{t a n y} d y L e t e^{x} - 1 = t \Rightarrow e^{x} = \frac{d t}{d x} \Rightarrow d x = \frac{d t}{e^{x}} a n d t a n y = v \Rightarrow s e c^{2} y = \frac{d v}{d y} \Rightarrow d y = \frac{d v}{s e c^{2} y} ∴ \int \frac{e^{x}}{t} \frac{d t}{e^{x}} = \int \frac{s e c^{2} y}{v} \frac{d v}{s e c^{2} y} \Rightarrow l o g | t | = l o g | v | - l o g | C | \Rightarrow l o g | e^{x} - 1 | = l o g | t a n y | - l o g | C | \Rightarrow l o g | C (e^{x} - 1) | = l o g | t a n y | [∵ l o g m + l o g n = l o g m n] \Rightarrow C (e^{x} - 1) = t a n Y (∵ l o g m = l o g n \Rightarrow m = n)$
which is the required general solution.

Suggest Corrections

Similar questions

e^{x} tan y d x + (1 - e^{x}) {sec}^{2} y d y = 0

For the differential equation in given question find the general solution.

$y l o g y d x - x d y = 0$

For the differential equation in given question find the general solution.

$(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0$

For each of the differential equation in given question find the general solution.
$s e c^{2} x t a n y d x + s e c^{2} y t a n x d y = 0$

For the differential equation in given question find the general solution.

$\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$

Join BYJU'S Learning Program