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Question

# The solution of the differential equation $\frac{dy}{dx}=\frac{x-y+3}{2\left(x-y\right)+5}$ is

A

$2\left(x-y\right)+\mathrm{log}\left(x-y\right)=x+c$

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B

$2\left(x-y\right)-\mathrm{log}\left(x-y+2\right)=x+c$

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C

$2\left(x-y\right)+\mathrm{log}\left(x-y+2\right)=x+c$

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D

None of these

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Solution

## The correct option is C $2\left(x-y\right)+\mathrm{log}\left(x-y+2\right)=x+c$Explanation for the correct optionGiven: $\frac{dy}{dx}=\frac{x-y+3}{2\left(x-y\right)+5}$Let $v=x-y$$\frac{dv}{dx}=1-\frac{dy}{dx}$Now ,$\frac{dy}{dx}=\frac{x-y+3}{2\left(x-y\right)+5}\phantom{\rule{0ex}{0ex}}⇒1-\frac{dv}{dx}=\frac{v+3}{2v+5}\phantom{\rule{0ex}{0ex}}⇒\frac{dv}{dx}=1-\frac{v+3}{2v+5}\phantom{\rule{0ex}{0ex}}⇒\frac{dv}{dx}=\frac{2v+5-v-3}{2v+5}\phantom{\rule{0ex}{0ex}}⇒\frac{dv}{dx}=\frac{v+2}{2v+5}\phantom{\rule{0ex}{0ex}}⇒\frac{dv}{dx}=\frac{v+2}{2v+5}\phantom{\rule{0ex}{0ex}}⇒\frac{2v+5}{v+2}dv=dx$Integrating both sides$\int \frac{2v+5}{v+2}dv=\int dx\phantom{\rule{0ex}{0ex}}⇒\int 2+\frac{1}{v+2}dv=\int dx\phantom{\rule{0ex}{0ex}}⇒2v+\mathrm{log}\left(v+2\right)=x+C$Put $v=x-y$Therefore $2\left(x-y\right)+\mathrm{log}\left(x-y+2\right)=x+C$Hence option C is correct.

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