What is the general solution of the differential equation ? xdy/dx=2/x+2-y given that x=-1,y=0

1 Answer
Nov 9, 2017

y = (ln|x| + x + 1)/x

Explanation:

We have:

xdy/dx=2/x+2-y ..... [A]

We can rearrange [A] as follows:

dy/dx=2/x^2 + 2/x - y/x

dy/dx + y/x = 2/x^2 + 2/x ..... [B]

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

dy/dx + P(x)y=Q(x)

So we form an Integrating Factor;

I = e^(int P(x) dx)
\ \ = exp( int \ 1/x \ dx)
\ \ = exp( lnx )
\ \ = x

And if we multiply the DE [B] by this Integrating Factor, I, we will have a perfect product differential;

\ \ x \ dy/dx + y = 2/x+2x

:. d/dx( x y ) = 2/x+2

Which we can directly integrate to get:

xy = int \ 2/x+2x \ dx
\ \ \ \ = 2ln|x| + 2x + C

Using the initial Condition, x=-1,y=0 we have:

0 = 2ln|1| - 2 + C => C = 2

Leading to the General Solution:

xy = 2ln|x| + 2x + 2
=> y = (ln|x| + x + 1)/x