calculus
posted by sammy .
consider the differential equation dy/dx= (y  1)/ x squared where x not = 0
a) find the particular solution y= f(x) to the differential equation with the initial condition f(2)=0
(b)for the particular solution y = F(x) described in part (a) find lim F(x) X goes to infinity

dy/dx = (y  1)/ x^2
Use the integration method of separation of variables.
dy/(y1) = dx/x^2
ln (y1) = 1/x + C
You say that the initial condition is
f(x) = y = 0 at x =2
Are you sure the initial condition is not f(0) = 2 ? You cannot have a logarithm of a negative number. 
Start like above, but once you gget to the logarithm part, it is in absolute values, so it is ln(1) which is 0. Solve from there

The two statements above is wrong.
you first move around the
dy/dx = (y1)/x^2 into
dy/(y1) = dx/x^2
Then you integral where it turns to
ln( (y1)/C ) = 1/x
Next you move the e from the ln to the other side
(y1)/C = e^(1/x)
After that you times C to both sides and move the 1 afterwords
y1 = Ce^(1/x) > y = Ce^(1/x)+1
I not sure how you do part (b) though.
Sorry