What is the general solution of the differential equation ? xdy/dx=2/x+2-y given that x=-1,y=0
1 Answer
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,
\ \ 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,
0 = 2ln|1| - 2 + C => C = 2
Leading to the General Solution:
xy = 2ln|x| + 2x + 2
=> y = (ln|x| + x + 1)/x