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

We have:

Rearranging [A] can be done as follows:

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

Thus, an Integrating Factor is formed.

which we can integrate directly to obtain:

Getting to the Overall Solution:

To find the general solution of the given differential equation (x \frac{{dy}}{{dx}} = \frac{2}{x} + 2 - y), we can follow these steps:

1. First, rewrite the equation in standard form: [x \frac{{dy}}{{dx}} + y = \frac{2}{x} + 2]

2. This equation is a linear first-order ordinary differential equation. To solve it, we can use an integrating factor. The integrating factor, denoted by (I(x)), is defined as: [I(x) = e^{\int P(x)dx}] where (P(x)) is the coefficient of (y) in the standard form equation.

3. In our case, (P(x) = \frac{1}{x}), so the integrating factor becomes: [I(x) = e^{\int \frac{1}{x}dx} = e^{\ln|x|} = |x|]

4. Multiply both sides of the standard form equation by the integrating factor: [|x| \cdot \left(x \frac{{dy}}{{dx}} + y\right) = |x| \cdot \left(\frac{2}{x} + 2\right)]

5. After simplifying, we get: [|x| \cdot \frac{{dy}}{{dx}} + |x| \cdot y = 2 + 2|x|]

6. Notice that (|x|) behaves differently for positive and negative values of (x). We'll handle these cases separately.

7. For (x > 0), the equation becomes: [x \frac{{dy}}{{dx}} + xy = 2 + 2x]

8. This is now a separable differential equation. We can rearrange terms and integrate both sides with respect to (x) to solve for (y).

9. For (x < 0), the equation becomes: [-x \frac{{dy}}{{dx}} - xy = 2 - 2x]

10. This equation is also separable. Rearrange terms and integrate both sides with respect to (x) to solve for (y).

11. Once we find the solutions for both cases, we can combine them using appropriate constants of integration to form the general solution.

12. After obtaining the general solution, we can use the initial condition (x = -1, y = 0) to find the particular solution by substituting these values into the general solution and solving for the constants of integration.