Separable differential equations

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

\[y' - f(x)g(y) = 0\]

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

\[y' = xy\].

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

Next, suppose \(y \neq 0\). 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

\[y = C e^{\frac{1}{2}x^2}\tag{Eq:egode}\]

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,

\[v(t) = \frac{mg}{k}\left(1 - e^{-\frac{k}{m}t}\right).\]

□

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 \(y' = u + u' x\) 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

\[y' = \frac{xy}{x^2 + y^2}.\]

This can be rearranged to

\[y' = \frac{\frac{y}{x}}{1 + \left(\frac{y}{x}\right)^2}.\]

Thus, it is homogeneous. Let \(u = \frac{y}{x}\), and we have \(y'= u + xu'\) so the given ODE becomes

\[u + xu' = \frac{u}{1 + u^2}\]

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.) □

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