$\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{x}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ + y = tan x
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ + $\mathit{s}\mathit{e}{\mathit{c}}^2$ x.y = $\mathit{s}\mathit{e}{\mathit{c}}^2$ x tan x
This equation is in the form of $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ + py = Q
here p = $\mathit{s}\mathit{e}{\mathit{c}}^2$ x and Q = $\mathit{s}\mathit{e}{\mathit{c}}^2$ x tan x
Integrating Factor , I.F = ${\mathit{e}}^{\int \mathit{p}\mathit{d}\mathit{x}}$ = ${\mathit{e}}^{\int \mathit{s}\mathit{e}{\mathit{c}}^2\mathit{x}\mathit{d}\mathit{x}}$ = ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$
The general solution can be given by
y (I.F.) = ∫ (Q × I.F.) dx + C ... (1)
Let tanx = t
⇒ $\frac{\mathit{d}}{\mathit{d}\mathit{x}}$ (tan x) = $\frac{\mathit{d}\mathit{t}}{\mathit{d}\mathit{x}}$
⇒ $\mathit{s}\mathit{e}{\mathit{c}}^2$ x = $\frac{\mathit{d}\mathit{t}}{\mathit{d}\mathit{x}}$
⇒ $\mathit{s}\mathit{e}{\mathit{c}}^2$ x dx = dt
Therefore, equation 1 becomes :
$\mathit{y}.{\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ = ∫ $\left({\mathit{e}}^{\mathit{t}}.\mathit{t}\right)$ dt
⇒ y . ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ = ∫ $\left({\mathit{e}}^{\mathit{t}}.\mathit{t}\right)$ dt + C
⇒ y . ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ = t . ∫ ${\mathit{e}}^{\mathit{t}}$ dt - ∫ $\left(\frac{\mathit{d}}{\mathit{d}\mathit{t}}\left(\mathit{t}\right).\int {\mathit{e}}^{\mathit{t}}\right)\mathit{d}\mathit{t}$ + C
⇒ y . ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ = $\mathit{t}.{\mathit{e}}^{\mathit{t}}-\int {\mathit{e}}^{\mathit{t}}\mathit{d}\mathit{t}$ + C
⇒ y . ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ = $\mathit{t}.{\mathit{e}}^{\mathit{t}}-{\mathit{e}}^{\mathit{t}}$ + C
⇒ y . ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ = (t - 1) ${\mathit{e}}^{\mathit{t}}$ + C
⇒ y . ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ = (tan x - 1) ${\mathit{e}}^{\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ + C
⇒ y = (tan x - 1) + $\mathit{C}{\mathit{e}}^{-\mathit{t}\mathit{a}\mathit{n}\mathit{x}}$ , where C is an arbitary constant