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

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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