Exercises

1. Solve each of the following initial value problems.

   `(dP)/(dt)=2P, Ptext[(]0text[)]=-1`	`(dP)/(dt)=2P^3, Ptext[(]0text[)]=0.5`
   `(dP)/(dt)=2P, Ptext[(]0text[)]=0`	`(dP)/(dt)=2P^3, Ptext[(]0text[)]=1`
   `(dP)/(dt)=2P^2, Ptext[(]0text[)]=2`	`(dP)/(dt)=2P^3, Ptext[(]0text[)]=-1`
   `(dP)/(dt)=2P^2, Ptext[(]0text[)]=-1`	`(dP)/(dt)=2P^3, Ptext[(]0text[)]=0`

2. Solve each of the following initial value problems.

   `(dy)/(dt)=1/t, ytext[(]1text[)]=0`	`(dy)/(dt)=1/y, ytext[(]0text[)]=1`
   `(dy)/(dt)=1/t, ytext[(]1text[)]=1`	`(dy)/(dt)=1/y, ytext[(]0text[)]=-1`

3. For each of the following initial value problems, find the largest interval on which your solution from Exercise 2 is valid.

   `(dy)/(dt)=1/t, ytext[(]1text[)]=0`	`(dy)/(dt)=1/y, ytext[(]0text[)]=1`
   `(dy)/(dt)=1/t, ytext[(]1text[)]=1`	`(dy)/(dt)=1/y, ytext[(]0text[)]=-1`

4. Solve each of the following initial value problems.

   `(dy)/(dt)=e^(-y), ytext[(]1text[)]=0`	`(dy)/(dt)=e^(-t), ytext[(]0text[)]=1`
   `(dy)/(dt)=e^y, ytext[(]-1text[)]=0`	`(dy)/(dt)=e^t, ytext[(]0text[)]=1`

5. For each of the following initial value problems, find the largest interval on which your solution from Exercise 4 is valid.

   `(dy)/(dt)=e^(-y), ytext[(]1text[)]=0`	`(dy)/(dt)=e^(-t), ytext[(]0text[)]=1`
   `(dy)/(dt)=e^y, ytext[(]-1text[)]=0`	`(dy)/(dt)=e^t, ytext[(]0text[)]=1`

6. Solve each of the following initial value problems.

   `(dy)/(dt)=e^(-y//2), ytext[(]1text[)]=0`	`(dy)/(dt)=e^(-y//2), ytext[(]2text[)]=0`
   `(dy)/(dt)=e^(y//2), ytext[(]-1text[)]=0`	`(dy)/(dt)=e^(y//2), ytext[(]-2text[)]=0`

7. For each of the following initial value problems, find the largest interval on which your solution from Exercise 6 is valid.

   `(dy)/(dt)=e^(-y//2), ytext[(]1text[)]=0`	`(dy)/(dt)=e^(-y//2), ytext[(]2text[)]=0`
   `(dy)/(dt)=e^(y//2), ytext[(]-1text[)]=0`	`(dy)/(dt)=e^(y//2), ytext[(]-2text[)]=0`

8. The initial value problem in Activity 4,

   `(dP)/(dt)=kP^2,   Ptext[(]0text[)]=P_0,`

   is another population model of some interest; we will return to it later. In what important ways does the solution of this problem differ from the solution, `Ptext[(]t text[)]=P_0 e^(kt)`, to the apparently similar initial value problem,

   `(dP)/(dt)=kP,   Ptext[(]0text[)]=P_0`?

9. In Section 5.1, we modeled the acceleration of a falling body by the initial value problem

   `(dv)/(dt)=g-kv,   vtext[(]0text[)]=0.`

   By use of a scaling argument (i.e., changing the scale on the `v`-axis), we saw that the unique solution of this initial value problem is

   `v=g/k(1-e^(-kt))=g/k - g/k e^(-kt)`.

   1. Find this solution another way, by separating the variables in `(dv)/(dt)=g-kv`.
   2. Write the left-hand side of `v=g/k - g/k e^(-kt)` as `(ds)/(dt)`, and find position `s` (distance fallen) as an explicit function of `t`.