Separable Differential Equations

A separable differential equation is one in which the first derivative \( y' \) of the unknown function can be expressed as the product of two functions: one depending only on \( x \), and the other only on \( y \): $$ y' = f(x) \cdot g(y) $$

To solve such equations, follow these steps:

1. Separate the variables, assuming \( y \ne 0 \). Move all terms involving \( y \) to one side and those involving \( x \) to the other: $$ \frac{dy}{dx} = f(x) \cdot g(y) $$ $$ \frac{dy}{g(y)} = f(x)\, dx $$
2. Integrate both sides with respect to their respective variables: $$ \int \frac{dy}{g(y)} = \int f(x)\, dx $$

This yields the general solution \( y(x) \), valid for \( y \ne 0 \).

Any constant solutions, such as \( y = 0 \), should be considered separately.

What are constant solutions? Constant (or trivial) solutions correspond to values of \( y \) for which \( y' = 0 \). Since \( y' = f(x)g(y) \), then \( y' = 0 \) implies either \( f(x) = 0 \) or \( g(y) = 0 \). To find constant solutions, set \( g(y) = 0 \) and solve for \( y \). If such values exist, they must be included in the complete solution set.

Examples

Example 1

Consider the differential equation:

$$ y' = 2xy^2 $$

This is a separable differential equation, since it can be expressed as a product of a function of \( x \) and a function of \( y \): \( f(x) = 2x \), \( g(y) = y^2 \).

Rewriting \( y' \) in Leibniz notation:

$$ \frac{dy}{dx} = 2x y^2 $$

We now separate the variables:

$$ \frac{dy}{y^2} = 2x\, dx $$

Integrating both sides:

$$ \int y^{-2}\, dy = \int 2x\, dx $$

Computing the antiderivatives:

$$ -\frac{1}{y} = x^2 + c $$

Solving for \( y \):

$$ y = -\frac{1}{x^2 + c} $$

This is the general solution of the differential equation.

We must also consider the constant solution \( y(x) = 0 \):

$$ y(x) = 0 $$

Note. The constant (or trivial) solution arises because \( g(y) = y^2 \) vanishes when \( y = 0 \). In this case, $$ y' = 2x \cdot 0^2 = 0 $$ so the equation is satisfied identically.

The complete solution set is therefore:

$$ y(x) = -\frac{1}{x^2 + c} \quad \text{or} \quad y(x) = 0 $$

Example 2

Solve the equation: $$ y' = \frac{\cos x}{\cos y} $$

We identify the functions: \( f(x) = \cos x \), \( g(y) = 1 / \cos y \).

In differential form: $$ \frac{dy}{dx} = \frac{\cos x}{\cos y} $$

Separate the variables: $$ \cos y\, dy = \cos x\, dx $$

Integrate both sides: $$ \int \cos y\, dy = \int \cos x\, dx $$

Evaluating the integrals: $$ \sin y = \sin x + c $$

To isolate \( y \), apply the inverse sine function: $$ y = \arcsin(\sin x + c) $$

Therefore, the general solution is: $$ y(x) = \arcsin(\sin x + c) $$

Note. In this case, there are no constant solutions, because \( g(y) = 1/\cos y \) is never equal to zero. Thus, the expression \( y' = \frac{\cos x}{\cos y} \) never vanishes due to \( g(y) \).

Example 3

Consider the differential equation:

$$ y' = \frac{x+1}{3y^2} $$

This is a separable equation and can be solved by rearranging the variables.

First, express the derivative in differential form and isolate the variables:

$$ \frac{dy}{dx} = \frac{x+1}{3y^2} $$

Multiplying both sides accordingly:

$$ 3y^2\, dy = (x+1)\, dx $$

Now integrate both sides with respect to their corresponding variables:

$$ \int 3y^2\, dy = \int (x+1)\, dx $$

Computing the antiderivatives yields:

$$ y^3 = \frac{x^2}{2} + x + c $$

Solving for \( y \), we take the cube root of both sides:

$$ y = \sqrt[3]{\frac{x^2}{2} + x + c} $$

This represents the general solution of the differential equation.

Note. As in previous cases, no constant solution arises here because the function \( g(y) = \frac{1}{3y^2} \) never vanishes for any real \( y \). Therefore, no constant value of \( y \) can satisfy the equation.

Example 4

Now consider the equation:

$$ y' = \sin(x) \cdot e^y $$

This equation is separable. We identify: \( f(x) = \sin x \), \( g(y) = e^y \).

Rewriting in differential form:

$$ \frac{dy}{dx} = \sin(x) \cdot e^y $$

Separating the variables:

$$ \frac{dy}{e^y} = \sin(x)\, dx \quad \Rightarrow \quad e^{-y}\, dy = \sin(x)\, dx $$

Integrating both sides gives:

$$ \int e^{-y}\, dy = \int \sin(x)\, dx $$

Evaluating the integrals:

$$ -e^{-y} = -\cos(x) + c $$

Multiplying both sides by -1 for clarity:

$$ e^{-y} = \cos(x) + c $$

Taking the natural logarithm of both sides:

$$ -y = \ln[\cos(x) + c] $$

Thus, the general solution is:

$$ y(x) = -\ln[\cos(x) + c] $$

Note. A constant solution does not exist in this case, as the exponential function \( g(y) = e^y \) is strictly positive for all real \( y \). Consequently, \( g(y) = 0 \) has no real solution.

Further Remarks

1. Separable differential equations are first-order equations expressed in standard form.
2. Autonomous differential equations of the form $$ u' = g(u) $$ are a particular case of separable equations, since they can be rewritten as $$ u' = 1 \cdot g(u) $$ with \( f(t) = 1 \) implicitly present.

And so forth.