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/(y-1) = dx/x^2

ln (y-1) = -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

To find the particular solution to the given differential equation with the initial condition, we can use separation of variables. Here's how you can do it:

a) Find the particular solution y = f(x) with the initial condition f(2) = 0:

1. Start with the given differential equation: dy/dx = (y - 1) / x^2.

2. Rearrange the equation by multiplying both sides by x^2 to separate the variables: x^2 dy = (y - 1) dx.

3. Integrate both sides with respect to x:

∫x^2 dy = ∫(y - 1) dx.

4. Evaluate the integrals:

(1/3) x^3 + C1 = y - x + C2,

where C1 and C2 are integration constants.

5. Combine the integration constants as a single constant:

(1/3) x^3 + C = y - x,

where C = C1 + C2.

6. Apply the initial condition f(2) = 0:

(1/3) (2)^3 + C = 0 - 2,

(8/3) + C = -2.

7. Solve for C:

C = -2 - (8/3),

C = -22/3.

8. Substitute the value of C into the equation:

(1/3) x^3 - (22/3) = y - x.

9. Rearrange the equation to solve for y:

y = (1/3) x^3 - x - (22/3).

Therefore, the particular solution to the differential equation with the initial condition f(2) = 0 is y = f(x) = (1/3) x^3 - x - (22/3).

b) To find lim F(x) as x approaches infinity:

To evaluate lim F(x) as x approaches infinity, we need to determine the behavior of F(x) as x gets larger and larger.

Looking at the equation y = f(x) = (1/3) x^3 - x - (22/3), as x approaches infinity, the dominant term becomes (1/3) x^3.

Therefore, lim F(x) as x approaches infinity is infinity.

The two statements above is wrong.

you first move around the
dy/dx = (y-1)/x^2 into

dy/(y-1) = dx/x^2

Then you integral where it turns to

ln( (y-1)/C ) = -1/x

Next you move the e from the ln to the other side

(y-1)/C = e^(-1/x)

After that you times C to both sides and move the -1 afterwords

y-1 = Ce^(-1/x) --> y = Ce^(-1/x)+1

I not sure how you do part (b) though.
Sorry