Integrating Factor

The Method of Integrating Factors

Let 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 p} and 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 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:

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 \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 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 \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. 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 \blacksquare}

Notice that from proposition 1 that integrating factors 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)} are not unique. In fact, there are infinitely many integrating factors. This can be see when evaluating the indefinite integral in 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}} which will result in getting 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^{P(t) + C}} where 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 P} is any antiderivative if 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 p} and where 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 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 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 \frac{dy}{dt} + \frac{2y}{t} = \frac{\sin t}{t^2}} .

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

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 \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 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 C} as a constant:

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 \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 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 \frac{dy}{dt} - \frac{y}{t} + te^{-t} = 0} .

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

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 \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 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 C} as a constant:

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 \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}}