Separable ODEs

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Section 15.3 Separable ODEs

Separable ODEs are first order ODEs with the special form:

\begin{equation} \frac{dy}{dx}=h(x)g(y).\tag{15.3.1} \end{equation}

where \(h(x)\) and \(g(y)\) are arbitrary functions, i.e. the \(x\) and \(y\) dependencies are separated into the product of two functions. A solution will exist wherever these functions are sufficiently smooth and \(y\) has no values for which \(g(y)\) is zero.

Method.

First, write (15.3.1) as a differentials equation by multiplying through by \(dx\text{.}\) Then divide by \(g(y)\) so that all the \(y\)-dependence is on one side of the equation and all the \(x\)-dependence is on the other side of the equation. Finally, integrate both sides of the equation.

\begin{align} dy\amp =h(x)g(y)\, dx,\tag{15.3.2}\\ \frac{1}{g(y)}\, dy\amp =h(x)\, dx,\tag{15.3.3}\\ \int\frac{1}{g(y)}\, dy\amp =\int h(x)\, dx.\tag{15.3.4} \end{align}

You may or may not be able to solve the resulting implicit equation for the dependent variable, in this case \(y\text{.}\)

If you do the integral as an indefinite integral, don't forget to add an arbitrary constant. (Adding the constant to one side of the equation is sufficient.) If you know an initial or boundary condition, the integrals become definite integrals and you can use this information to specify the value of the unknown constant.

Activity 15.2.

Solve the differential equation

\begin{equation} \frac{dy}{dx}=y\left(3x-2\right).\tag{15.3.5} \end{equation}

\begin{equation} y =e^{\left(\frac{3x^2}{2}-2x+C\right)} \tag{15.3.6} \end{equation}

Following the method above, we have

\begin{align} dy\amp =\left(3x-2\right)\, dx,\tag{15.3.7}\\ \frac{1}{y}\, dy\amp =\left(3x-2\right)\, dx,\tag{15.3.8}\\ \int\frac{1}{y}\, dy\amp =\int \left(3x-2\right)\, dx.\tag{15.3.9}\\ \ln y\amp =\frac{3x^2}{2}-2x+C.\tag{15.3.10} \end{align}

If we are given an initial condition such as \(y(0)=7\text{,}\) then we can use this to find the value \(C=\ln 7\text{.}\) The final answer is

\begin{align} \ln y\amp =\frac{3x^2}{2}-2x+\ln 7,\tag{15.3.11}\\ y\amp =e^{\left(\frac{3x^2}{2}-2x+\ln 7\right)},\tag{15.3.12}\\ \amp =7e^{\left(\frac{3x^2}{2}-2x\right)}.\tag{15.3.13} \end{align}