Exercises

The first three questions require a yes or no answer while the last requires a more lengthy solution

1. Is this a linear first order ODE?
   $\frac{df}{dt}+\cos \left(\lambda f\right)=0$

   NO because the unknown function appears in a cosine function which is not a linear function.

2. Is this a linear first order ODE?
   $\frac{df}{dt}+\cos \left(\lambda t\right)f=0$

   YES because the unknown function f(t) appears only to the first power.
   $p\left(s\right)=\cos \left(\lambda t\right),g\left(s\right)=0$

3. Is this a linear first order ODE?
   $\frac{df}{dt}-\cos \left(\alpha t\right)=f$

   YES because the unknown function f(t) appears only to the first power. But be careful to first rearrange the ODE in our notation in the discussion. Thus $p\left(s\right)=-1,g\left(s\right)=\cos \left(\alpha t\right)$

4. What is the solution of the following linear first order ODE?
   $\frac{df}{dt}=7-3f,f\left(0\right)=5$

   Solution: Put the equation in our preferred notation and identify p(s) and g(s).
   $\frac{df}{dt}+3f=7,f\left(0\right)=5,p\left(s\right)=3,g\left(s\right)=7$
   $\begin{array}{l}u\left(t\right)=\exp \left[{\displaystyle {\int }_{t}p\left(s\right)ds}\right],p\left(s\right)=3\\ u\left(t\right)=\exp \left[{\displaystyle {\int }_{t}3ds}\right]=\exp \left[3t\right]\\ \frac{1}{u\left(t\right)}=\exp \left[-3t\right]\end{array}$
   Next
   ${\displaystyle {\int }_{t}u\left(s\right)g\left(s\right)ds}+c={\displaystyle {\int }_t\exp \left[3t\right]\left(7\right)ds}+c=\frac{7}{3}\exp \left[3t\right]+c$
   Putting this together
   $f\left(t\right)=\exp \left[-3t\right]\left\{\frac{7}{3}\exp \left[3t\right]+c\right\}$ or $f\left(t\right)=\left\{\frac{7}{3}+c\exp \left[-3t\right]\right\}$
   Now use the initial condition to find the constant c.
   $f\left(0\right)=5,f\left(0\right)=\left\{\frac{7}{3}+c\exp \left[0\right]\right\}=5$
   $\frac{7}{3}+c=5\Rightarrow c=\frac{8}{3}$ . Thus $f\left(t\right)=\frac{1}{3}\left\{7+8\exp \left[-3t\right]\right\}$