`(x+y)dx - xdy = 0` Solve the first-order differential equation by any appropriate method

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kseddy123 eNotes educator| Certified Educator

Given` (x+y)dx - xdy= 0`

=>` (x+y) - xdy/dx= 0`

=>` x+y-xy'=0`

=>` x+y=xy'`

=> `1+y/x=y'`

=> `y' -y/x=1`

when the first order linear ordinary differential 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=1--------(1)`

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

on comparing both we get,

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

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) *(1) dx +c)/e^(int (-1/x) dx)`

first we shall solve

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

so

proceeding further, we get

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

 =`(int (1/x) *(1) dx +c)/(1/x)`

=`(ln(x) +c)/(1/x ) = x(ln(x)+c)`

So , `y(x) = x(ln(x)+c)`