If dy/dx = (x + y + 2)/(x - y) and y(0) = 2, find y(2)

The correct answer is 0.

Analysis of the Image & Problem Setup:
Based on the solution steps shown in the image, the differential equation being solved is:

$$\frac{dy}{dx}=\frac{x+y-2}{x-y}$$

This is a non-homogeneous first-order ordinary differential equation. To make it homogeneous, we shift the origin to the intersection point of the lines $x+y-2=0$ and $x-y=0$, which intersect at $(1,1)$.
Thus, we use the substitutions:

$$x=X+1,y=Y+1$$

This gives $dx=dX$ and $dy=dY$. The differential equation transforms to:

$$\frac{dY}{dX}=\frac{X+Y}{X-Y}$$

Step 1: Homogeneous Substitution
Let $Y=vX$. Differentiating both sides with respect to $X$ gives:

$$\frac{dY}{dX}=v+X\frac{dv}{dX}$$

Substitute this into the transformed equation:

$$v+X\frac{dv}{dX}=\frac{1+v}{1-v}$$

Step 2: Separating Variables
Subtract $v$ from both sides:

$$X\frac{dv}{dX}=\frac{1+v}{1-v}-v=\frac{1+v-v+v^2}{1-v}=\frac{1+v^2}{1-v}$$

Rearranging to separate variables gives:

$$\frac{(1-v)dv}{1+v^2}=\frac{dX}{X}$$

Step 3: Integration
Integrating both sides:

$$\int \frac{1}{1+v^2}dv-\int \frac{v}{1+v^2}dv=\int \frac{dX}{X}$$

Evaluating the integrals:

$${\tan }^{-1}\left(v\right)-\ln \left(\sqrt{1+v^2}\right)=\ln \left|X\right|+c$$

Step 4: Re-substitution
Substitute back $v=\frac{Y}{X}=\frac{y-1,}{x-1}$ and $X=x-1$:

$${\tan }^{-1}\left(\frac{y-1}{x-1}\right)-\ln \sqrt{1+{\left(\frac{y-1}{x-1}\right)}^2}=\ln |x-1|+c$$

Step 5: Apply the Initial Condition
We are given $y\left(0\right)=2$, so we substitute $x=0$ and $y=2$:

$${\tan }^{-1}\left(\frac{2-1}{0-1}\right)-\ln \sqrt{1+{\left(\frac{2-1}{0-1}\right)}^2}=\ln |0-1|+c$$

Simplify the terms:

$${\tan }^{-1}(-1)-\ln \sqrt{2}=\ln \left(1\right)+c$$

Since ${\tan }^{-1}(-1)=-\frac{\pi }{4}$ and $\ln \left(1\right)=0$, we find the constant $c$:

$$c=-\frac{\pi }{4}-\ln \sqrt{2}$$

Step 6: Finding y(2)
Now we set $x=2$ to solve for $y$:

$${\tan }^{-1}\left(\frac{y-1}{2-1}\right)-\ln \sqrt{1+{\left(\frac{y-1}{2-1}\right)}^2}=\ln |2-1|+c$$

Simplifying the terms:

$${\tan }^{-1}(y-1)-\ln \sqrt{1+{(y-1)}^2}=-\frac{\pi }{4}-\ln \sqrt{2}$$

Comparing the left-hand side and right-hand side directly:
If we substitute $y=0$:

$$\mathrm{LHS}={\tan }^{-1}(-1)-\ln \sqrt{1+{(-1)}^2}=-\frac{\pi }{4}-\ln \sqrt{2}=\mathrm{RHS}$$

Thus, $y=0$ satisfies the equation.
Therefore, $y\left(2\right)=0$.