Differential Equations — Class 12 Maths Solution

.: We have, ${e^x}\tan ydx + (1 - {e^x}){\sec ^2}ydy = 0$

$\Rightarrow {e^x}\tan ydx = - (1 - {e^x}){\sec ^2}ydy$

$\Rightarrow \cfrac{{{e^x}}}{{1 - {e^x}}}dx = \cfrac{{ - {{\sec }^2}y}}{{\tan y}}dy$

…(1)
Integrating (1) both sides,

we get
$\int {\cfrac{{{e^x}}}{{1 - {e^x}}}} dx = - \int {\cfrac{{{{\sec }^2}y}}{{\tan y}}} dy$

Let ${I_1} = \int {\cfrac{{{e^x}}}{{1 - {e^x}}}} dx$ and ${I_2} = \int {\cfrac{{{{\sec }^2}y}}{{\tan y}}} dy$

Now, ${I_1} = \int {\cfrac{{{e^x}}}{{1 - {e^x}}}} dx$

Putting $1 - {e^x} = t \Rightarrow - {e^x}dx = dt$

$\Rightarrow {I_1} = \int {\cfrac{{ - dt}}{t}} = - \log |t| + \log {C_1} = - \log |1 - {e^x}| + \log {C_1}$

Now, ${I_2} = \int {\cfrac{{{{\sec }^2}y}}{{\tan y}}} dy$

Putting $\tan y = t \Rightarrow {\sec ^2}ydy = dt$

$\Rightarrow {I_2} = \int {\cfrac{{dt}}{t}} = \log |t| + \log {C_2} = \log |\tan y| + \log {C_2}$

Also, ${I_1} = - {I_2}$
$\Rightarrow$ $- \log ({C_1}(1 - {e^x})) = - \log ({C_2}\tan y) \Rightarrow {C_1}(1 - {e^x}) = {C_2}\tan y$

$\Rightarrow \tan y = C(1 - {e^x})$,

which is required solution. $\left[ {C = \cfrac{{{C_1}}}{{{C_2}}}} \right]$