Question
Solve the following differential equation: $\cos^{2} x \frac{dy}{dx} + y = \tan x$
$\frac{\text{dy}}{\text{dx}} + \sec^{2} \text{x y} = \tan \text{x}.\sec^{2}\text{x}$
$\text{I.F} = \text{e}^{\int\text{pdx}}=\text{e}^{\int\sec^2\text{x dx}}=e^{\tan\text{x}}$
$\therefore$
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}} $
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$\text{e}^{-\text{y}}\sec^2\text{y dy}=\text{dx}+\text{x dy}$
$\text{y}(1+\text{e}^{\text{x}})\text{dy}=(\text{y}+1)\text{e}^{\text{x}}\text{ dx}$
| Tablets | Iron | Calcium | Vitamin |
| X | 6 | 3 | 2 |
| Y | 2 | 3 | 4 |
$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$