Separable differential equations

Separable differential equations are the simplest differential equations. They are of the following form:

$$y^{\prime }-f(x)g(y)=0$$

where $f(x)$ is a function of $x$ and $g(y)$ is a function of $y$. This equation can be formally rearranged into

$$\frac{dy}{g(y)}=f(x)dx$$

which can be integrated as

$$\int \frac{dy}{g(y)}=\int f(x)dx.$$

This method of solving differential equations is called the separation of variables.

Example. Let us solve the differential equation

$$y^{\prime }=xy$$.

If $y=0$ (constant), the given differential equation is clearly satisfied. Thus, $y=0$ is a solution.

Next, suppose $y\ne 0$. By separating variables, we have

$$\frac{dy}{y}=xdx.$$

Integrating both sides,

$$\log y=\int \frac{dy}{y}=\int xdx=\frac{1}{2}{x}^2+c$$

where $c$ is a constant. Thus, we have

$$\begin{array}{}\text{(Eq:egode)}& y=C{e}^{\frac{1}{2}{x}^2}\end{array}$$

where we set $C=e^c$. But this solution includes the case when $y=0$ (constant) if we set $C=0$. Thus, the general solution is given by (Eq:egode). □

Remark. The solution as in (Eq:egode) is called a general solution since it includes all possible solutions. In contrast, a solution like $y=e^{\frac{1}{2}{x}^2}$ is called a special solution as it is a special case where $C=1$. □

Example. Consider a body with mass $m$ falling in the air under the gravity of Earth. Let $v=v(t)$ be the velocity of the body at time $t$. The air resistance is given by $kv$. Then the equation of motion is given by

$$m\frac{dv}{dt}=mg-kv$$

where $g$ is the gravity of Earth. Given the initial condition $v(0)=0$, let us solve this differential equation. By separating the variables, we have

$$\frac{dv}{mg-kv}=\frac{dt}{m}.$$

Integrating both sides gives

$$\log |mg-kv|=-\frac{kt}{m}+C$$

where $C$ is a constant. By the initial condition $v(0)=0$, we have

$$C=\log (mg).$$

Thus,

$$v(t)=\frac{mg}{k}(1-e^{-\frac{k}{m}t}).$$

□

In the above example, we solved a differential equation that satisfied a given initial condition. Such a problem is called an initial value problem.

Homogeneous differential equations

A differential equation of the form

$$\frac{dy}{dx}=f\left(\frac{y}{x}\right)$$

is said to be homogeneous. In this case, if we let $u=\frac{y}{x}$, or $y=ux$, we have $y^{\prime }=u+u^{\prime }x$ so that

$$\frac{du}{dx}=\frac{\frac{dy}{dx}-u}{x}=\frac{f(u)-u}{x}.$$

Since this is separable, we can find $u=u(x)$, and thereby $y=y(x)=u(x)x$.

Example. Let's solve

$$y^{\prime }=\frac{xy}{x^2+y^2}.$$

This can be rearranged to

$$y^{\prime }=\frac{\frac{y}{x}}{1+{\left(\frac{y}{x}\right)}^2}.$$

Thus, it is homogeneous. Let $u=\frac{y}{x}$, and we have $y^{\prime }=u+x{u}^{\prime }$ so the given ODE becomes

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

which is rearranged to

$$-\frac{1+u^2}{u^3}du=\frac{dx}{x}.$$

Integrating both sides, we have

$$\frac{1}{2u^2}-\log |u|=\log |x|+C.$$

This results in

$$\frac{x^2}{2y^2}=\log |y|+C.$$

(The solution $y=y(x)$ is given as an implicit function.) □