3. What is a differential equation? An equation that resulted from differentiating another equation. dy dt = - 9.8t differential equation ex. v(t) = - 9.8t

4. What is a differential equation? An equation that resulted from differentiating another equation. This equation came from differentiating the position function: y = - 4.9t 2 + C ex. v(t) = - 9.8t

5. separation of variables comes from getting similar variables on one side of equation dy dt = - 9.8t dy dt = - 9.8t dy dt = - 9.8t dt * * dt both sides have d/dx attached, so can take antiderivative to get a new equation. Watch labels!!!!! Label is currently: meters/second

6. dy dt = - 9.8t take antiderivative to get a new equation. Watch labels!!!!! anti-derivative: y = - 4.9 t 2 + C Label is now meters. This is the solution to the differential equation!!!

7. 2001 FR Question #6 6.(b) Find y = f(x) by solving the differential equation = y 2 ( 6 - 2x ) with the initial condition f(3) =. dy dx 1414 1) separate variables 2) take anti-derivative 3) Isolate the solution in terms of C 3) find C… if dropping absolute value from ln  note j = ± C to alleviate the issue of ln y = - value. 4) rewrite original equation.

8. The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year. Write a differential equation that expresses this fact. Solve it to find an equation that expresses population as a function of time. Let P = population t years after the present. =.05P dP dt *label is people/yr Rate of change of the population

9. The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year. Write a differential equation that expresses this fact. Solve it to find an equation that expresses population as a function of time. P = 1000 e 0.05 t *with this equation, however P = 1051.27 when t = 1. P should equal 1050. Must change up a little!!!! start over. To check, find the population after one year given …..” The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year” ; Use the equation we found to check if when t=1 the population comes out right! 4

10. The population of the little town of Scorpion Gulch is now 1000 people. The population is presently growing at about 5% per year. Write a differential equation that expresses this fact. Solve it to find an equation that expresses population as a function of time. *when t = 1, P should equal 1050. (1000*.05= 50 plus original 1000) *When t = 0, P = 1000 = kP dP dt We must find a new constant (k) since 0.05 didn’t work!!!

11. *when t = 1, P should equal 1050. (1000*.05= 50 plus original 1000) *When t = 0, P = 1000 = kP dP dt *Separate variables in = kP dP dt dP P = k dt *find antiderivative of each side ln | P | + C = k t + C *solve for P with algebra!!

12. Now have: |P| = Ce k t Solve for C using: *You can solve for C as soon as you find antiderivative if there is not an absolute value involved……otherwise solve with C intact AND note  j = ± C !!! *When t = 0, P = 1000 Solve for k (what we wanted) using above equation and *when t = 1, P = 1050.  P = j e k t

13. therefore: P = 1000e ln t In general: y = y o e k t where y o is original value at time, feet, or whatever = 0 21 20