Differential Equations and Seperation of Variables

1. Differential Equations and Seperation of Variables

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

3. What is a differential equation? An equation that resulted from differentiating another equation. ex. v(t) = - 9.8t This equation came from differentiating the position function: y = - 4.9t2 + C

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

5. dy = - 9.8t dt take antiderivative to get a new equation. Watch labels!!!!! anti-derivative: y = - 4.9 t2 + C Label is now meters. This is the solution to the differential equation!!!

6. dy dx 1 4 2001 FR Question #6 6.(b) Find y = f(x) by solving the differential equation = y2 ( 6 - 2x ) with the initial condition f(3) = . 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.

7. dP dt = .05P 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. Rate of change of the population Let P = population t years after the present. *label is people/yr

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. .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! P = 1000 e0.05 t *with this equation, however P = 1051.27 when t = 1. P should equal 1050. 4 Must change up a little!!!! start over

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. = kP dP dt We must find a new constant (k) since 0.05 didn’t work!!! *When t = 0, P = 1000 *when t = 1, P should equal 1050. (1000*.05= 50 plus original 1000)

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

11. Solve for k (what we wanted) using above equation and *when t = 1, P = 1050. 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  P = j e k t

12. therefore: P = 1000e ln t 21 20 In general: y = yo e k t where yo is original value at time, feet, or whatever = 0 Law of exponential change!!