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Question

The solution of the differential equation $$\displaystyle \frac{dy}{dx}=\frac{x+y}{x}$$ satisfying the condition $$y(1)=1$$ is:


A
y=xlogx+x2
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B
y=xe(x1)
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C
y=xlogx+x
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D
x+logx
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Solution

The correct option is C $$y=x\log x+x$$
$$\displaystyle\frac { dy }{ dx } =\frac { x+y }{ x }$$
$$\displaystyle\Rightarrow xdy=xdx+ydx$$
$$\displaystyle\Rightarrow xdy-ydx=xdx$$

$$\displaystyle\Rightarrow \frac { xdy-ydx }{ { x }^{ 2 } } =\frac { x }{ { x }^{ 2 } } dx$$

$$\displaystyle\Rightarrow d\left( \frac { y }{ x }  \right) =\frac { 1 }{ x } dx$$
Integrating both sides, we get 
$$\displaystyle\frac { y }{ x } =\ln { x } +c$$
For $$y\left( 1 \right) =1\Rightarrow c=1$$
$$\therefore y=x\ln { x } +x$$

Mathematics

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