#### Solving with an Integrating Factor 1. Put equation in form $\frac{dy}{dt}+P(t)=f(t)$ 2. Find the Integrating Factor $\mu(t)=e^{\int P(t)dt}$ 3. Multiply both sides of the equation by the Integrating Factor - $\mu(t)\frac{dy}{dt}+P(t)\mu(t)y=\mu(t)f(t)$ 4. By the product rule the left side is the derivative of $\mu(t)y$ - $\frac{dy}{dt}[\mu(t)y] =\mu(t)f(t)$ 5. Integrate both sides - $\int \frac{dy}{dt}[\mu(t)y]dt =\int \mu(t)f(t)dt$ - Integral and derivative on left side cancel 6. Rearrange as necessary to isolate $y(t)$ #### Examples with Integrating Factors - All examples were scanned and typset into LaTex [[AI LaTex Typsetting|using Claude]]! ##### 1. $\frac{dy}{dt} - \frac{5}{t}y = t^5$ $ \begin{align} \frac{dy}{dt} - \frac{5}{t}y &= t^5 \\ P(t) &= \frac{-5}{t} \\\qquad \qquad \mu(t) &= e^{\int \frac{-5}{t}dt} = e^{-5\ln t} = t^{-5} \\ \\ t^{-5}\frac{dy}{dt} - \frac{5}{t}t^{-5}y &= t^5 t^{-5} \\[0.5cm] t^{-5}\frac{dy}{dt} - 5t^{-6}y &= 1 \\[0.5cm] \frac{d}{dt}\left[t^{-5}y\right] &= 1 \\[0.5cm] \int\frac{d}{dt}t^{-5}y\,dt &= \int 1\,dt \\[0.5cm] t^{-5}y &= t + c \\[0.5cm] y &= \frac{t}{t^{-5}} + \frac{c}{t^{-5}} \\\\ y(t) &= t^6 + ct^5 \end{align} $ --- ##### 2. $\frac{dy}{dt} = -2ty + 4e^{-t^2}$ $ \begin{align} \frac{dy}{dt} = -2ty + 4e^{-t^2} \\ P(t) &= 2t \\ \mu(t) &= e^{\int 2t\,dt} = e^{t^2} \\\\ \frac{dy}{dt} + 2ty = 4e^{-t^2} \\[0.5cm] e^{t^2}\frac{dy}{dt} + 2te^{t^2}y &= 4e^{-t^2}e^{t^2} \\[0.5cm] \frac{d}{dt}\left[e^{t^2}y\right] &= 4 \\[0.5cm] \int\frac{d}{dt}e^{t^2}y\,dt &= \int 4\,dt \\[0.5cm] e^{t^2}y &= 4t + C \\[0.5cm] y &= 4te^{-t^2} + Ce^{-t^2} \\[0.5cm] y(t) &= e^{-t^2}(4t + C) \end{align} $ --- ##### 3. $\frac{dy}{dt} - \frac{2y}{t} = t^3e^t$ $ \begin{align} \frac{dy}{dt} - \frac{2y}{t} &= t^3e^t \\[0.5cm] P(t) &= -\frac{2}{t} \\ \mu(t) &= e^{\int -\frac{2}{t}dt} = e^{-2\ln t} = t^{-2} \\[0.5cm] t^{-2}\frac{dy}{dt} - \frac{2}{t}t^{-2}y &= t^3e^t t^{-2} \\[0.5cm] \frac{d}{dt}\left[t^{-2}y\right] &= te^t \\[0.5cm] \int\frac{d}{dt}\left[t^{-2}y\right]dt &= \int te^tdt \\[0.5cm] t^{-2}y &= \int te^t dt \qquad \text{Integrate by parts} \\ \qquad u &= t \quad dv=e^t \\ \qquad du &= q \quad v=e^t \\&\int uv = uv - \int vdu \\&= te^t - \int e^t dt \\&= te^t - e^t + C \\[0.5cm] t^{-2}y &= te^t - e^t + C \\[0.5cm] y &= \frac{te^t}{t^{-2}} - \frac{e^t}{t^{-2}} + \frac{C}{t^{-2}} \\[0.5cm] &= t^3e^t - t^2e^t + Ct^2 \\[0.5cm] y(t) &= t^3e^t - t^2e^t + Ct^2 \end{align} $