Example Solve the following initial value problem． $y^{\prime} = \frac{x - y}{x + y}, y(0) = 1$

Answer Functions $M(x,y)$ and $N(x,y)$ are 1st degree homogeneous functions. Then we divide the right-hand side by $x$.

$$y^{\prime} = \frac{1 - (y/x)}{1 + (y/x)} = f(\frac{y}{x})$$

Now we let $y = vx$. Then we have

$$v^{\prime}x + v = \frac{1 - v}{1 + v}$$

From this, we obtain

$$v^{\prime}x = \frac{1 - v}{1 + v} - v = \frac{1 - v - v - v^{2}}{1 + v}$$

or

$$\frac{1 + v}{v^{2} + 2v - 1}v^{\prime} = -\frac{1}{x}$$

Integrate both sides with respect to $x$.

$$\frac{1}{2}\log{\vert v^{2} + 2v - 1\vert} = -\log{(cx)},$$

Here, $c$ is a constant.

$$(v^{2} + 2v - 1)x^{2} = c_{1}.$$

Note that $c_{1} =1/c^{2}$. Then it is a constant．Lastly, set $v = y/x$ and simplify, we have

$$y^{2} + 2xy - x^{2} = c_{1}$$

Now using the initial condition $y(0) = 1$, we have

$$y^{2} + 2xy - x^{2} = 1$$