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Section 9.3 Separable ODEs

Separable ODEs are first order ODEs with the special form:

\begin{equation} \frac{dy}{dx}=h(x)g(y).\label{eqn-separable}\tag{9.3.1} \end{equation}

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


First, write (9.3.1) as a differentials equation by multiplying through by \(dx\text{.}\) Then divide by \(h(x)\) 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{9.3.2}\\ \frac{1}{g(y)}\, dy\amp =h(x)\, dx,\tag{9.3.3}\\ \int\frac{1}{g(y)}\, dy\amp =\int h(x)\, dx.\label{eqn-dsep}\tag{9.3.4} \end{align}
Activity 9.3.1.

Solve the differential equation

\begin{equation} \frac{dy}{dx}=y\left(3x-2\right).\label{eqn-sepex}\tag{9.3.5} \end{equation}
\begin{equation} y =7e^{\left(\frac{3x^2}{2}-2x\right)} \tag{9.3.6} \end{equation}

Following the method above, we have

\begin{align} dy\amp =\left(3x-2\right)\, dx,\tag{9.3.7}\\ \frac{1}{y}\, dy\amp =\left(3x-2\right)\, dx,\tag{9.3.8}\\ \int\frac{1}{y}\, dy\amp =\int \left(3x-2\right)\, dx.\tag{9.3.9}\\ \ln y\amp =\frac{3x^2}{2}-2x+C.\tag{9.3.10} \end{align}

Since we are doing an indefinite integral, it is important for us to remember, in the second to last step, to add the unspecified constant \(C\) to one side of the equation. 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{9.3.11}\\ y\amp =e^{\left(\frac{3x^2}{2}-2x+\ln 7\right)},\tag{9.3.12}\\ \amp =7e^{\left(\frac{3x^2}{2}-2x\right)}.\tag{9.3.13} \end{align}