Question

e^xtanydx+(1-e^x)sec²ydy=0, solve the following differential equation

Answer 1

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\rm \: {e}^{x} \: tany \: dx + (1 - {e}^{x}) \: {sec}^{2}y \: dy \: = \: 0$

can be rewritten as

$\rm \: {e}^{x} \: tany \: dx = ({e}^{x} - 1) \: {sec}^{2}y \: dy \:$

can be further rewritten as

$\rm \: \dfrac{{e}^{x}}{{e}^{x} - 1} dx \: = \: \dfrac{ {sec}^{2}y}{tany} dy$

On integrating both sides, we get

$\rm \: \displaystyle\int\rm \dfrac{{e}^{x}}{{e}^{x} - 1} dx \: = \:\displaystyle\int\rm \dfrac{ {sec}^{2}y}{tany} dy$

Now,

$\rm \: \dfrac{d}{dx}({e}^{x} - 1) = {e}^{x}$

and

$\rm \: \dfrac{d}{dy}tany = {sec}^{2}y$

Also, we know that,

$\boxed{ \rm{ \:\displaystyle\int\rm \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c \: \: }}$

So, above expression can be rewritten as

$\rm \: log |{e}^{x} - 1| = log |tany| + logc$

$\rm \: log |{e}^{x} - 1| = log |c(tany)|$

$\rm \: {e}^{x} - 1 = c \: tany$

Hence, Solution of differential equation is

$\rm\implies \:\rm \:\boxed{ \rm{ \: {e}^{x} - 1 = c \: tany \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$