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  • Integrating factor of the differential equation cosxdy/dx + ysinx = 1 is: A. cosx B. tanx C. sec x D. sinx

Integrating factor of the differential equation \cos x \frac{d y}{d x}+y \sin x=1 is:
A. cosx
B. tanx
C. sec x
D. sinx

Answers (1)

Differential equation is
$$ \\ \cos x \frac{d y}{d x}+y \sin x=1 \\ \Rightarrow \frac{d y}{d x}+\frac{y \sin x}{\cos x}=\frac{1}{\cos x} \\ \Rightarrow \frac{d y}{d x}+(\tan x) y=\sec x $$
Compare
$$ \frac{d y}{d x}+(\tan x) y=\sec x $$
With

\frac{d y}{d x}+P y=Q^{\prime}$ we get, $P=\tan x$ and $Q=\sec x$
The IF integrating factor is given by 
$$ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{\sin \mathrm{x}}{\cos \mathrm{x}} \mathrm{dx}} $$
Substitute \cos x=t$  hence
\\\frac{\mathrm{dt}}{\mathrm{dx}}=-\sin \mathrm{x}$ \\$\sin x d x=-d t$
Resubstitute the value of t
\Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log (\cos \mathrm{x})^{-1}}$

$$\\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log (\cos \mathrm{x})^{-1}} \\ \Rightarrow \mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\log {\sec \mathrm{x}}} = \sec x$$

hence IF is sec x

Option C is correct.

Posted by

infoexpert22

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