The given differential equation is:
${\mathit{e}}^{\mathit{x}}$ tan y dx + $\left(1-{\mathit{e}}^{\mathit{x}}\right)\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{y}$ dy = 0
⇒ ${\mathit{e}}^{\mathit{x}}$ tan y dx = - $\left(1-{\mathit{e}}^{\mathit{x}}\right)\mathit{s}\mathit{e}{\mathit{c}}^{\mathit{y}}$ dy
⇒ ${\mathit{e}}^{\mathit{x}}$ tan y dx = $\left({\mathit{e}}^{\mathit{x}}-1\right)\mathit{s}\mathit{e}{\mathit{c}}^{\mathit{y}}$ dy
⇒ ${\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{\mathit{x}}-1$ dx = $\frac{\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{y}}{\mathit{t}\mathit{a}\mathit{n}\mathit{y}}$ dy
On integrating on both sides, we get
∫ ${\mathit{e}}^{\mathit{x}}-{\mathit{e}}^{\mathit{x}}-1$ dx = ∫ $\frac{\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{y}}{\mathit{t}\mathit{a}\mathit{n}\mathit{y}}$ dy ... (1)
Let ${\mathit{I}}_1$ = ∫ $\frac{\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{y}}{\mathit{t}\mathit{a}\mathit{n}\mathit{y}}$ dy
Put tany = t
⇒ $\mathit{s}\mathit{e}{\mathit{c}}^2$ y dy = t
∴ ∫ $\frac{\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{y}}{\mathit{t}\mathit{a}\mathit{n}\mathit{y}}$ dy = ∫ $\frac{\mathit{d}\mathit{t}}{\mathit{t}}$ = log |t| = log tan y ... (2)
Let ${\mathit{I}}_2$ = ∫ $\frac{{\mathit{e}}^{\mathit{x}}}{{\mathit{e}}^{\mathit{x}}-1}$ dx
Put ${\mathit{e}}^{\mathit{x}}$ - 1 = u
∴ ${\mathit{e}}^{\mathit{x}}$ dx = du
∫ $\frac{{\mathit{e}}^{\mathit{x}}}{{\mathit{e}}^{\mathit{x}}-1}$ dx = ∫ $\frac{\mathit{d}\mathit{u}}{\mathit{u}}$
= log u
= log $\left({\mathit{e}}^{\mathit{x}}-1\right)$ ... (3)
From i , ii and iii , we get
log tan y = log (${\mathit{e}}^{\mathit{x}}$ - 1) + log C
⇒ log tan y = log C (${\mathit{e}}^{\mathit{x}}$ - 1)
⇒ tan y = C (${\mathit{e}}^{\mathit{x}}$ - 1)
The solution of the given differential equation is tan y = C (${\mathit{e}}^{\mathit{x}}$ - 1).