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Section 4.3 Separable Equations

References.

Prelude.

We have seen that the special kind of differential equation

\begin{equation} \frac{dy}{dx} = f(x)\tag{4.3.1} \end{equation}

can be solved by simply integrating; the general solution is given by the indefinite integral

\begin{equation} y = \int f(x)\, dx.\tag{4.3.2} \end{equation}

A variant of this can be applied with the right-hand side instead depends only on the unknown, \(y\)

\begin{equation} \frac{dy}{dx} = g(y)\tag{4.3.3} \end{equation}

One way to do this is to divide through by the right-hand side and then integrate:

\begin{equation*} \int \frac{1}{g(y)}\frac{dy}{dx} dx = \int 1\, dx. \end{equation*}

The left hand-side is as for substitution, with \(y\) in the role of \(u\text{;}\) that is, we can effectively cancel the two occurences of \(dx\text{,}\) getting

\begin{equation} \int \frac{1}{g(y)} dy = x + C\tag{4.3.4} \end{equation}

A complication is that this then gives an equation involving \(y\) rather than a direct formula for \(y\text{,}\) so one might have to do some equation solving.

Find the general solution of

\begin{equation} \frac{dy}{dx} = y.\tag{4.3.5} \end{equation}
Solution.

The procedure above gives

\begin{equation*} \int \frac{1}{y}\frac{dy}{dx} dx = \int \frac{1}{y} dy = \int 1\, dx \end{equation*}

so that

\begin{equation*} \ln |y| = x + C. \end{equation*}

Solving for \(|y|\) first gives \(|y| = e^{x+C} = e^x e^C\) so the two options for \(y\) are \(y = (\pm e^C)e^x\)

The factor \(k = \pm e^C\) can be any non-zero constant, so we have the solutions

\begin{equation} y = k e^x,\; k \neq 0.\tag{4.3.6} \end{equation}

In fact, \(k=0\) also works, giving \(y=0\text{;}\) we lost that possibility along way because dividing by \(y\) was not possible in that special case.

Separation of Variables.

The two special cases above can be combined into a more general form

\begin{equation} \frac{dy}{dx} = f(x) g(y)\tag{4.3.7} \end{equation}

called separable because the two variables can be separated into an integral in \(x\) on one side and an integral in \(y\) on the other.

To do this, start much as with (4.3.3): divide by \(g(y)\) to get

\begin{equation} \frac{1}{g(y)}\frac{dy}{dx} = f(x)\tag{4.3.8} \end{equation}

and then integrate and use the substitution rule as above to “cancel” the occurences of \(dx\) at left:

\begin{equation} \int \frac{1}{g(y)}\frac{dy}{dx} dx = \int \frac{1}{g(y)} dy = \int f(x)\, dx\tag{4.3.9} \end{equation}

Integrating each side gives an equation connecting \(y\) to \(x\text{;}\) as above, you might need to solve this equation for \(y\) as a function of \(x\text{;}\) however, sometimes this equation connecting the two variables is enough information!

Find the general solution of

\begin{equation} \frac{dy}{dx} = -\frac{x}{y}.\tag{4.3.10} \end{equation}
Solution.

Multiplying by \(y\) on both sides gives

\begin{equation*} y \frac{dy}{dx} = -x \end{equation*}

and integrating,

\begin{equation*} \int y \frac{dy}{dx} dx = \int y\, dy = -\int x\, dx \end{equation*}

so that \(y^2/2 = -x^2/2 + C\) or

\begin{equation*} x^2 + y^2 = 2C. \end{equation*}

The solution is always (part of) a circle, which is often all one needs to know.

Solving for \(y\) explicitly in terms of \(x\text{,}\) being careful with the two possibilities for the square root, one gets

\begin{equation*} y = \pm\sqrt{2C- x^2} \end{equation*}

so to be precise, the solution is either the top half or the bottom half of a circle.

Solve the initial value problem for the above differential equation (4.3.10) with \(y(x_0) = y_0\text{;}\) that is, find the solution that passes through the point \((x_0, y_0)\text{.}\)

The above example shows an interesting possibility with first order differential equations: they can sometimes be viewed as describing the slope \(dy/dx\) at each point \((x, y)\) on a curve (like a circle), with the solution being given as an equation for the curve, and maybe the solution curve is not describable explicitly as the graph of a function \(y = F(x)\text{.}\)

Applications of Separation of Variables.

We omit this, but read this subsection in Openstax Calculus if you are interested in some applications to physics and physical chemistry, or just see a few more worked examples.

Study Guide.

Study Calculus Volume 2, Section 4.1 2 ; in particular

  • Problem-Solving Strategy,

  • Example 10 and Checkpoint 10,

  • and one or several exercises from each of the ranges 123–132 and 133–142.

https://openstax.org/books/calculus-volume-2/pages/4-3-separable-equations
https://openstax.org/books/calculus-volume-2/pages/4-1-basics-of-differential-equations