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Section 11.1 Exact First Order DEs

Solve the differential equation

\begin{equation} x \dfrac{dy}{dx} + y = \sin(x)\label{Eq1_example_1}\tag{11.1.1} \end{equation}
Answer.

\(y(x) = -\dfrac{\cos(x)}{x} + \dfrac{C}{x}\)

Solution.

This \(1^{\text{st}}\) order differential equation is not separable, i.e. it cannot be put into the form

\begin{equation*} \dfrac{dy}{dx} = f(x)g(y)\text{.} \end{equation*}

However it is easily solved once we observe that, for a function \(y=y(x)\) (and using the product rule)

\begin{equation*} \dfrac{d}{dx} (xy) = x \dfrac{dy}{dx} + y\text{.} \end{equation*}

Using this observation, equation (11.1.1) becomes

\begin{equation*} \dfrac{d}{dx} (xy) = \sin(x) \end{equation*}

and hence, on integrating both sides with respect to \(x\text{,}\)

\begin{equation*} xy = -\cos(x) + C\text{.} \end{equation*}

Thus the general solution to (11.1.1) is

\begin{equation*} y(x) = -\dfrac{\cos(x)}{x} + \dfrac{C}{x}\text{.} \end{equation*}

The DE in Example 11.1.1 is an example of another class of DEs for which we can find a general solution.

Definition 11.1.2. Exact First Order DE.

A \(1^{\text{st}}\) order differential equation of the form

\begin{equation*} \dfrac{d}{dx} \left( F(x,y) \right) = f(x) \end{equation*}

is called an exact first order DE.

To solve an exact DE we just need to integrate both sides of the equation with respect to \(x\text{.}\) Sometimes a DE that is not exact can be rearranged so that it is exact.

Rearrange the following DE so that it is exact

\begin{equation} \dfrac{dy}{dx}+\dfrac{y}{x} = \dfrac{\sin(x)}{x}\text{.}\label{Eq2_example_2}\tag{11.1.2} \end{equation}
Answer.

\(x\dfrac{dy}{dx}+y = \sin(x)\)

Solution.

By comparing (11.1.2) with (11.1.1) we can see that (11.1.2) can be made exact by multiplying both sides by \(x\text{.}\)

If a first order DE can be made exact by multiplying both sides by a function \(f\text{,}\) then \(f\) is called an integrating factor for the DE.

Show that \(f(x)=x^4\) is an integrating factor for the DE

\begin{equation*} \dfrac{dy}{dx} + \dfrac{4}{x} y = x+1\text{.} \end{equation*}
Solution.

If \(f(x) = x^4\) is an integrating factor for the DE then multiplying both sides of the equation by \(f\) will produce an exact DE. Now

\begin{align*} x^4 \left( \dfrac{dy}{dx} + \dfrac{4}{x} y \right) \amp = x^4(x+1)\\ x^4 \dfrac{dy}{dx} + 4x^3y \amp = x^5 + x^4\\ \dfrac{d}{dx} \left(x^4 y \right) \amp = x^5 + x^4 \end{align*}

which is exact.

Note: To solve this DE all we have to do now is integrate both sides with respect to \(x\text{.}\) On doing this we obtain

\begin{equation*} y(x) = \frac{1}{6} x^2 + \frac{1}{5} x + \frac{C}{x^4}\text{.} \end{equation*}

Unfortunately, not all first order DEs have an integrating factor. However for an important class of first order DEs, namely first order linear DEs, we can always find an integrating factor. We discuss this class of DEs in the next section.