Question

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

Answer 1

`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`