DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

$\frac{\text{dy}}{\text{dx}} + \sec^{2} \text{x y} = \tan \text{x}.\sec^{2}\text{x}$

$\text{I.F} = \text{e}^{\int\text{pdx}}=\text{e}^{\int\sec^2\text{x dx}}=e^{\tan\text{x}}$

$\therefore$ The solution is

$\text{y} .e^{\tan{\text{x}}} = \int e^{\tan\text{x}} . \tan\text{x}.\sec^{2}\text{x dx + c}$

$\text{Let} \tan{\text{x}} = \text{z} \Rightarrow \sec^{2}\text{x dx = dz}$

$\therefore \int e^{\tan \text{x}} \tan \text{x} \sec^{2}\text{x dx} = \int {\text{z e}^{\text{z}} } \text{dz + c} $

$= \text{z}.e^{\text{z}} - e^{\text{z}} + \text{c} = e^{\text{z}} (\text(z - 1) + \text{c}$

$\text{y e}^{\tan\text{x}} = e^{\tan\text{x}} ( \tan{\text{x}} - 1) + \text{c}e^{-\tan\text{x}} $