Separation of variables

The first order differential equation is expressed as

$\displaystyle y^{\prime} = f(x,y)$

$\displaystyle M(x,y)dx + N(x,y)dy = 0$

The latter equation is called the total differential equation. Now consider

$\displaystyle y^{\prime} = -\frac{y}{x}, \ xdx + ydy = 0$

Both equations are the same except at $x = 0$

. Thus we formally treat both equations as the same equation.

If the differential equation $y^{\prime} = f(x,y)$ is expressed in the form

$\displaystyle y^{\prime} = F(x)G(y),$

then we say the differential equation is separable. To solve this equation,

$\displaystyle y^{\prime}(x) = F(x)G(y(x))$

we express $y^{\prime}(x)$ as $\frac{dy}{dx}$ , and then write

$\displaystyle \frac{dy}{G(y)} = F(x) dx$

Now integrate both sides with respect to $x$

to obtain

$\displaystyle \int \frac{dy}{G(y)} dx = \int F(x)dx.$

Example 1..4 Find the general solution of $y^{\prime} = \frac{xy}{y - 1}$ .

SOLUTION We can rewrite this differential equation as

$\displaystyle \frac{(y-1)dy}{y} = x dx$

This is separable and integrate both sides to obtain

$\displaystyle y - \log\vert y\vert = \frac{x^2}{2} + c$

This is an implicit solution. Now let $c_1 = \log{c}$ . Then

$\displaystyle y - \log(c_1 y)^2 = x^2$

Note that $y(x) \equiv 0$ is a solution. But no matter how you choose $c_1$

, it is impossible to obtain $y = 0$

. Thus

is a singular solution. As we noted above, when we are asked to find the general solution, we solve the differential equation by quadrature. Thus do not worry about a singular solution.

Let $T_{1}$ and $T_{2}$ be the temparature of two objects facing each other. Then the heat transfers from warmer body to cooler body in the time $dt$ is given by $dQ$ and

$\displaystyle dQ = \kappa (T_{1} - T_{2})dt.\$ where $\displaystyle \ \kappa\$ is constant

Example 1..5 Let the iron ball be heated to $100^{\circ}c$ . Then submerge the iron ball into the water whose temparature is kept at $10^{\circ}c$ . After $4$