Separable Equations and Autonomous Equations

1Introduction

This lecture explains separation of variables not as a formal rearrangement of $$dy$$dy and $$dx$$dx, but as an application of the chain rule in reverse.

Separation applies when the rate of change of an unknown function factors as

with one factor depending only on $$x$$x and the other only on $$y$$y. This product structure permits the $$y$$y-dependent factor to be placed on one side and the $$x$$x-dependent factor on the other.

Division by $$g\left(y\right)$$g(y) excludes points at which $$g\left(y\right)=0$$g(y)=0. Therefore, constant solutions arising from such zeros must be checked before division. Some sources denote this factor by $$h\left(y\right)$$h(y); the same caution applies.

2Standard Form

A separable equation is a first-order differential equation that can be written as

An equation of the form

whose right-hand side does not explicitly contain the independent variable, is called an autonomous equation. It is separable by taking $$f\left(x\right)=1$$f(x)=1 and $$g\left(y\right)=F\left(y\right)$$g(y)=F(y).

3Proposition 1: An Implicit Relation Obtained by Separation Satisfies the Original Equation

Let $$I$$I and $$J$$J be intervals, let $$y:I\to J$$y:I\to J be differentiable, and suppose that $$g$$g is nonzero on $$J$$J. Consider

Assume that $$F^{\prime }\left(x\right)=f\left(x\right)$$F'(x)=f(x) on $$I$$I and $$G^{\prime }\left(y\right)=1/g\left(y\right)$$G'(y)=1/g(y) on $$J$$J. Then every such function $$y\left(x\right)$$y(x) satisfying

satisfies the original differential equation.

3.1Proof

Differentiate $$G\left(y\right(x\left)\right)=F\left(x\right)+C$$G(y(x))=F(x)+C with respect to $$x$$x. The chain rule gives

Using $$G^{\prime }\left(y\right)=1/g\left(y\right)$$G'(y)=1/g(y) and $$F^{\prime }\left(x\right)=f\left(x\right)$$F'(x)=f(x) gives

Since $$g\left(y\right)\ne 0$$g(y)\ne 0, multiplication by $$g\left(y\right)$$g(y) yields

Thus the implicit relation obtained by separation satisfies the original equation. This chain-rule calculation justifies the formal notation

4Solution Procedure

1. Rewrite the equation as $$y^{\prime }=f\left(x\right)g\left(y\right)$$y'=f(x)g(y).
2. First identify constant solutions satisfying $$g\left(y\right)=0$$g(y)=0.
3. On a range where $$g\left(y\right)\ne 0$$g(y)\ne 0, separate the variables:

1. Integrate both sides.
2. If conditions are prescribed, solve the resulting constraint equations for the arbitrary constants.
3. Verify the result by substitution into the original equation and all prescribed conditions.

5Proposition 2: Zeros of $$g$$g Give Constant Solutions

Suppose that

and that a constant $$a$$a satisfies $$g\left(a\right)=0$$g(a)=0. Then

is a constant solution.

5.1Proof

Since $$y\left(x\right)=a$$y(x)=a is constant, $$y^{\prime }=0$$y'=0. Also,

Therefore $$y^{\prime }=f\left(x\right)g\left(y\right)$$y'=f(x)g(y). This is why zeros of $$g$$g must be checked before division: the divided equation no longer represents these constant solutions.

6Example 1: $$y^{\prime }=xy$$y'=xy

The equation

is separable with $$f\left(x\right)=x$$f(x)=x and $$g\left(y\right)=y$$g(y)=y. First, $$g\left(0\right)=0$$g(0)=0 shows that $$y=0$$y=0 is a constant solution.

On a branch where $$y\ne 0$$y\ne 0,

Integration gives

Exponentiation gives

Since $$e^C>0$$e^C>0 and a constant sign can be absorbed into the arbitrary constant, the family can be written as

Allowing $$C=0$$C=0 includes the previously identified constant solution $$y=0$$y=0.

7Explicit and Implicit Solutions

An explicit solution represents the unknown directly as $$y=H\left(x\right)$$y=H(x).

An implicit solution represents the solution through a relation such as

The result of separation cannot always be solved explicitly for $$y$$y. In that case, a verified implicit relation still represents a solution on intervals where the implicit relation defines a differentiable function.

8Example 2: A Solution with Finite-Time Blow-Up

Consider

The constant function $$y=0$$y=0 is a solution of the differential equation but does not satisfy the initial condition. On the branch where $$y\ne 0$$y\ne 0,

Integration gives

The condition $$y\left(0\right)=1$$y(0)=1 yields $$C=-1$$C=-1, so

and therefore

This function is undefined at $$x=1$$x=1. Hence the maximal solution interval containing the initial point $$x=0$$x=0 is $$(-\mathrm{\infty },1)$$(-\infty,1). The example shows that a solution must be specified by both its formula and its domain.

9Autonomous Equations and Equilibria

An equilibrium solution is a constant solution whose value remains unchanged as the independent variable varies. For the autonomous equation

every zero $$F\left(a\right)=0$$F(a)=0 gives the equilibrium solution $$y=a$$y=a.

9.1Proof

If $$y=a$$y=a, then $$y^{\prime }=0$$y'=0. Since $$F\left(a\right)=0$$F(a)=0, the right-hand side is also zero; hence $$y^{\prime }=F\left(y\right)$$y'=F(y).

The sign of $$F\left(y\right)$$F(y) determines monotonicity before an explicit formula is obtained: $$F\left(y\right)>0$$F(y)>0 implies that $$y$$y increases, while $$F\left(y\right)<0$$F(y)<0 implies that $$y$$y decreases.

10Example 3: The Logistic Equation

The equation

is autonomous. Its equilibria are the zeros of $$F\left(y\right)=y(1-y)$$F(y)=y(1-y):

The sign table is

Thus, while a solution remains in $$0<y<1$$0<y<1, it increases, and the phase-line arrow points toward the equilibrium $$y=1$$y=1. This sign analysis determines local monotonicity before an explicit formula is derived. Convergence to $$y=1$$y=1 additionally requires verification that the solution remains in $$0<y<1$$0<y<1 and determination of its limit.

11A Nonseparable Example

The equation

contains $$x$$x and $$y$$y additively and cannot be factored as $$f\left(x\right)g\left(y\right)$$f(x)g(y). It must not be separated. Instead, rewrite it as

which is a first-order linear equation.

12Relation to Singular Solutions

Losing a constant solution with $$g\left(y\right)=0$$g(y)=0 through division and the existence of a singular solution are distinct issues. A singular solution satisfies the differential equation but cannot be obtained by selecting the arbitrary constants in its general solution family.

Nonuniqueness, in which multiple solutions satisfy the same initial condition, is a separate issue governed by the hypotheses of existence and uniqueness theorems; it is neither the definition nor a sufficient condition for a singular solution. After separation, verify every divided factor, the domain, and the prescribed initial conditions.

13Summary

For a separable equation, check the following four items.

1. Can it be written as $$y^{\prime }=f\left(x\right)g\left(y\right)$$y'=f(x)g(y)?
2. Do zeros of $$g\left(y\right)$$g(y) give constant solutions?
3. Does integration produce an explicit or implicit solution?
4. What is the domain of the resulting solution?

14Exercise