Examples:

1. Which of the following is a linear first order ODE?

   a. $\frac{df}{dt}+\lambda f^2=0$
   b. $\frac{df}{dt}+{\lambda }^{2}f=\ln \left(t\right)$
   c. $\frac{df}{dt}+{\lambda }^2{\left(f\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=7$
   

   a. $\frac{df}{dt}+\lambda f^2=0$ : NO, f appears to the second power so it is not linear
   b. $\frac{df}{dt}+{\lambda }^{2}f=\ln \left(t\right)$ : YES, This is a linear first order ODE
   c. $\frac{df}{dt}+{\lambda }^2{\left(f\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=7$ : NO, f is raised to the half power which makes it non-linear
   

2. What is the unknown function f(t) that solves the following linear first order ODE?

   $\frac{df}{dt}+2f=0,f\left(0\right)=10$
   Use the formula in the discussion to solve.

   Solution:
   $\begin{array}{l}u\left(t\right)=\exp \left[{\displaystyle {\int }_{t}p\left(s\right)ds}\right],p\left(s\right)=2\\ u\left(t\right)=\exp \left[{\displaystyle {\int }_{t}2ds}\right]=\exp \left[2t\right]\\ \frac{1}{u\left(t\right)}=\exp \left[-2t\right]\end{array}$
   Next
   ${\displaystyle {\int }_{t}u\left(s\right)g\left(s\right)ds}+c={\displaystyle {\int }_t\exp \left[2t\right]\left(0\right)ds}+c=0+c$
   thus
   $f\left(t\right)=c\exp \left[-2t\right]$ and $f\left(0\right)=10$ so the final result is $f\left(t\right)=10\exp \left[-2t\right]$ .

3. Solve the following linear first order ODE.

   $\frac{dy}{dt}=-\lambda y\left(t\right),y\left(0\right)=y_0$

   Solution: First, identify each of the terms in the equation with the formulas that you learned in the discussion. This ODE is written is a slightly different form than the standard ODE in the discussion. Thus we bring all the terms containing y(t) to the left hand side and put all the terms that contain just some other arbitrary function on the right hand side.

   $\frac{dy}{dt}+\lambda y\left(t\right)=0$ Now you can see $p\left(s\right)=\lambda$ and $g\left(s\right)=0$
   $\begin{array}{l}u\left(t\right)=\exp \left[{\displaystyle {\int }_{t}p\left(s\right)ds}\right],p\left(s\right)=\lambda \\ u\left(t\right)=\exp \left[{\displaystyle {\int }_t\lambda ds}\right]=\exp \left[\lambda t\right]\\ \frac{1}{u\left(t\right)}=\exp \left[-\lambda t\right]\end{array}$
   Next
   ${\displaystyle {\int }_{t}u\left(s\right)g\left(s\right)ds}+c={\displaystyle {\int }_t\exp \left[\lambda t\right]\left(0\right)ds}+c=0+c$
   and finally
   $y\left(t\right)=c\exp \left[-\lambda t\right]$ and $\text{ y}\left(0\right)=y_0$ so the final result is $y\left(t\right)=y_0\exp \left[-\lambda t\right]$