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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 $x$ .

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

Now we let $y = vx$

. Then we have

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

From this, we obtain

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

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

Integrate both sides with respect to $x$

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

Here,

is a constant.

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

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

and simplify, we have

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

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

, we have

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