Differential Equations — Class 12 Maths Solution

Given that, ${e^x}\cos ydx - {e^x}\sin ydy = 0$

$\Rightarrow$ ${e^x}\cos ydx = {e^x}\sin ydy$

$\Rightarrow$ $\frac{{dx}}{{dy}} = \tan y$

$\Rightarrow$ $dx = \tan ydy$

On integrating both sides,

we get $x = \log \sec y + C$

$\Rightarrow$ $x - C = \log \sec y$

$\Rightarrow$ $\sec y = {e^{x - c}}$

$\Rightarrow$ $\sec y = {e^x}{e^{ - c}}$

$\Rightarrow$ $\frac{1}{{\cos y}} = \frac{{{e^x}}}{{{e^c}}}$
$\Rightarrow$ ${e^x}\cos y = {e^c}$

$\Rightarrow$ ${e^x}\cos y = K$. [where, $K = {e^C}$]