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 $\mu (t)=e^{\int p(t)\;dt}$.

• 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:

${\begin{aligned}\quad {\frac {\mu '(t)}{\mu (t)}}=p(t)\\\quad {\frac {d}{dt}}\ln \mid \mu (t)\mid =p(t)\\\quad \int {\frac {d}{dt}}\ln \mid \mu (t)\mid \;dt=\int p(t)\;dt\\\quad \ln \mid \mu (t)\mid =\int p(t)\;dt\\\quad \mu (t)=\pm e^{\int p(t)\;dt}\end{aligned}}$

• 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 $\mu (t)=e^{\int p(t)\;dt}$ 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 $\mu (t)=e^{\int p(t)\;dt}$ 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:

${\begin{aligned}\quad \mu (t)=e^{\int p(t)\;dt}=e^{\int {\frac {2}{t}}\;dt}=e^{2\ln(t)}=e^{\ln(t^{2})}=t^{2}\end{aligned}}$

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

${\begin{aligned}\quad \mu (t){\frac {dy}{dt}}+\mu (t)p(t)y=\mu (t)g(t)\\\quad t^{2}{\frac {dy}{dt}}+2ty=\sin t\\\quad {\frac {d}{dt}}\left(t^{2}y\right)=\sin t\\\quad \int {\frac {d}{dt}}\left(t^{2}y\right)\;dt=\int \sin t\;dt\\\quad t^{2}y=-\cos t+C\\\quad y={\frac {-\cos t}{t^{2}}}+{\frac {C}{t^{2}}}\end{aligned}}$

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:

${\begin{aligned}\quad \mu (t)=e^{\int p(t)\;dt}=e^{\int -{\frac {1}{t}}\;dt}=e^{-\ln t}=e^{\ln \left({\frac {1}{t}}\right)}={\frac {1}{t}}\end{aligned}}$

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

${\begin{aligned}\quad \mu (t){\frac {dy}{dt}}+\mu (t)p(t)y=\mu (t)g(t)\\\quad {\frac {1}{t}}{\frac {dy}{dt}}-{\frac {1}{t^{2}}}y=-e^{-t}\\\quad {\frac {d}{dt}}\left({\frac {y}{t}}\right)=-e^{-t}\\\quad \int {\frac {d}{dt}}\left({\frac {y}{t}}\right)\;dt=-\int e^{-t}\;dt\\\quad {\frac {y}{t}}=e^{-t}+C\\\quad y=te^{-t}+tC\end{aligned}}$