Given` y'+y/x=0` and find the particular solution at y(2)=2

when the first order linear ordinary Differentian equation has the form of

`y'+p(x)y=q(x)`

then the general solution is ,

`y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)`

so,

`y'+y/x=0--------(1)`

`y'+p(x)y=q(x)---------(2)`

on comparing both we get,

`p(x) = 1/x and q(x)=0`

so on solving with the above general solution we get:

y(x)=`((int e^(int p(x) dx) *q(x)) dx +c)/ e^(int p(x) dx) `

=`((int e^(int (1/x) dx) *(0)) dx +c)/e^(int(1/x) dx)`

first we shall solve

`e^(int (1/x) dx)=e^(ln(x)) =x`

so

proceeding further, we get

y(x) =`((int e^(int (1/x) dx) *(0)) dx +c)/e^(int(1/x) dx)`

=`(0 +c)/(x)`

`=c/x`

so `y(x)=c/x`

now let us find the value of c at `y(2) =2 `

so,

`y(x)=c/x`

=> `2=c/2 `

=>` c=4`

so at y(2) =2 the particular solution is `y(x) =4/x`

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