Integrating factor of the differential equation $\cos\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\sin\text{x}=1$ is:
- $\cos\text{x}$
- $\tan\text{x}$
- $\sec\text{x}$
- $\sin\text{x}$
Solution:
we have, $\cos\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\sin\text{x}=1$
$\Rightarrow\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}\tan\text{x}=\sec\text{x}$
This is a linear differential equation.
On comparing it with $\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{P}\text{y}=\text{Q},$ we get
$\text{P}=\tan\text{x}$ and $\text{Q}=\sec\text{x}$
$\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\tan\text{xdx}}$
$=\text{e}^{\log\sec\text{x}}=\sec\text{x}$
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$(A)$ $5$ $(B)$ $7$ $(C)$ $\frac{-15}{2}$ $(D)$ $\frac{-17}{2}$
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