Separation of variables

The first order differential equation is expressed as

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

or

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

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

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

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

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

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

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

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

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

Now integrate both sides with respect to $x$ to obtain

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

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

SOLUTION We can rewrite this differential equation as

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

This is separable and integrate both sides to obtain

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

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

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

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 $y = 0$ 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

$$dQ = \kappa (T_{1} - T_{2})dt.\$$   where$$\ \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 $4$ minutes, the temparature of the iron ball is $60^{\circ}c$. Find the time when the temparature of the iron ball is $20^{\circ}c$.

SOLUTION We formulate this problem by using Newton's law of cooling. Then

$$\frac{dT}{dt} = - \kappa(T - 10), \ T(0) = 100, \ T(4) = 60,$$

where $T(t)$ is a temparature of the iron ball after $t$ minutes later. Since this differential equation is separable, we obtain

$$\frac{dT}{T-10} = -\kappa dt$$

Integrating both sides with respect to $t$. Then

$$\log\vert T-10\vert = -\kappa t + c$$

and

$$T(t) = ce^{-kt} + 10$$

Now by the boundary conditions, $T(0) = 100$ and $T(4) = 60$, we have

$$c = 90$$   and$$\ k = \frac{1}{4} \log{\frac{9}{5}} \doteq \frac{1}{4}(2.1972 - 1.6094) \doteq 0.147$$

Thus,

$$T(t) \doteq 90e^{{-0.147}t} + 10.$$

Now we find $t_0$ satisfying $T(t) = 20$.

$$20 \doteq 90e^{{-0.147} t_{0}} + 10$$

and

$$t_{0} \doteq \frac{\log{9}}{0.147} \doteq 14.9\ {\rm min} \ensuremath{\ \blacksquare}$$