Integrating Factor

The Method of Integrating Factors

Let $p$ and $g$ be functions of $t$ and consider the following first order differential equation:

${\begin{aligned}\quad {\frac {dy}{dt}}+p(t)y=g(t)\end{aligned}}$

If we multiply the both sides of the equation above by the function $\mu (t)$ we get that:

${\begin{aligned}\quad \mu (t){\frac {dy}{dt}}+\mu (t)p(t)y=\mu (t)g(t)\end{aligned}}$

If we can guarantee that $\mu '(t)=\mu (t)p(t)$, then notice that by applying the product rule for differentiation that we get: ${\frac {d}{dt}}\left(\mu (t)y\right)=\mu (t){\frac {dy}{dt}}+\mu '(t)y$ and substituting $\mu '(t)=\mu (t)y$ and we get that ${\frac {d}{dt}}\left(\mu (t)y\right)=\mu (t){\frac {dy}{dt}}+\mu (t)p(t)y$ which is exactly the lefthand side of the equation above. Thus we get that:

${\begin{aligned}{\frac {d}{dt}}\left(\mu (t)y\right)=\mu (t)g(t)\\\end{aligned}}$

The above differential equation can be solved by integrating both sides of the equation with respect to $t$ and isolating $y$. The question now arises on how we can find such a function $\mu (t)$.

Definition: If ${\frac {dy}{dt}}+p(t)y=g(t)$ is a first order differential equation, then $\mu (t)$ is called an Integrating Factor if for $\mu (t){\frac {dy}{dt}}+\mu (t)p(t)y=\mu (t)g(t)$ we have that $\mu '(t)=\mu (t)p(t)$.

The following proposition will give us a formula for obtaining the integrating factor for differential equations in the form ${\frac {dy}{dt}}+p(t)y=g(t)$.

Proposition 1: If ${\frac {dy}{dt}}+p(t)y=g(t)$ is a differential equation, then an integrating factor $\mu (t)$ of this equation is given by the formula Failed to parse (syntax error): {\displaystyle \mu (t) = e^{\int p(t) \: dt}} .

• Proof: We want to find $\mu (t)$ such that $\mu '(t)=\mu (t)p(t)$. We can rewrite this equation as as ${\frac {\mu '(t)}{\mu (t)}}=p(t)$ and then:

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \quad \frac{ \mu '(t)}{ \mu (t)} = p(t) \\ \quad \frac{d}{dt} \ln \mid \mu (t) \mid = p(t) \\ \quad \int \frac{d}{dt} \ln \mid \mu (t) \mid \: dt = \int p(t) \: dt \\ \quad \ln \mid \mu (t) \mid = \int p(t) \: dt \\ \quad \mu (t) = \pm e^{\int p(t) \: dt} \end{align}}

• Since we only need one integrating factor to solve differential equations in the form ${\frac {dy}{dt}}+p(t)y=g(t)$, we can more generally note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mu (t) = e^{\int p(t) \: dt}} is an integrating factor of this differential equation. $\blacksquare$

Notice that from proposition 1 that integrating factors $\mu (t)$ are not unique. In fact, there are infinitely many integrating factors. This can be see when evaluating the indefinite integral in Failed to parse (syntax error): {\displaystyle \mu (t) = e^{\int p(t) \: dt}} which will result in getting $\mu (t)=e^{P(t)+C}$ where $P$ is any antiderivative if $p$ and where $C$ is a constant. We will always use the simplest integrating factor in solving differential equations of this type.

Let's now look at some examples of applying the method of integrating factors.

Example 1

Find all solutions to the differential equation ${\frac {dy}{dt}}+{\frac {2y}{t}}={\frac {\sin t}{t^{2}}}$.

We first notice that our differential equation is in the appropriate form ${\frac {dy}{dt}}+p(t)y=g(t)$ where $p(t)={\frac {2}{t}}$ and $g(t)={\frac {\sin t}{t^{2}}}$. We compute our integrating factor as:

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \quad \mu (t) = e^{\int p(t) \: dt} = e^{\int \frac{2}{t} \: dt} = e^{2 \ln (t)} = e^{\ln (t^2)} = t^2 \end{align}}

Thus we have that for $C$ as a constant:

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad t^2 \frac{dy}{dt} + 2t y = \sin t \\ \quad \frac{d}{dt} \left ( t^2 y \right ) = \sin t \\ \quad \int \frac{d}{dt} \left ( t^2 y \right ) \: dt = \int \sin t \: dt \\ \quad t^2 y = -\cos t + C \\ \quad y = \frac{-\cos t}{t^2} + \frac{C}{t^2} \end{align}}

Example 2

Find all solutions to the differential equation ${\frac {dy}{dt}}-{\frac {y}{t}}+te^{-t}=0$.

We first rewrite our differential equation as ${\frac {dy}{dt}}-{\frac {y}{t}}=-te^{-t}$. We note that in this form we have $p(t)=-{\frac {1}{t}}$ and $g(t)=-te^{-t}$. We now find an integrating factor:

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \quad \mu (t) = e^{\int p(t) \: dt} = e^{\int -\frac{1}{t} \: dt} = e^{-\ln t} = e^{\ln \left ( \frac{1}{t} \right)} = \frac{1}{t} \end{align}}

Thus we have that for $C$ as a constant:

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \quad \mu (t) \frac{dy}{dt} + \mu (t) p(t) y = \mu (t) g(t) \\ \quad \frac{1}{t} \frac{dy}{dt} - \frac{1}{t^2} y = -e^{-t} \\ \quad \frac{d}{dt} \left ( \frac{y}{t} \right ) = -e^{-t} \\ \quad \int \frac{d}{dt} \left ( \frac{y}{t} \right ) \: dt = -\int e^{-t} \: dt \\ \quad \frac{y}{t} = e^{-t} + C \\ \quad y = te^{-t} + tC \end{align}}