1. By Holum Kwok

2. In order to prepare for the AP Calc AB Exam… Solve differential equations and use Dif EQs in modeling Find specific antiderivatives using initial conditions Study equation y’=ky and exponential growth

3. So Assuming you know how to integrate a function…

4. Separation of Variables Use to solve Dif EQs Ex. Find equation of a solution curve for dy/dx= -x/y 1. Move all x to one side and all y to one side: y dy= -x dx 2. Antidifferentiate both sides: ∫ y dy= ∫ -x dx y 2 /2 = -x 2 /2 +c (remember to plus C- you idiot!) 3. Isolating y: y 2 = -2x 2 /2 +2c (2C is a new constant, let’s call it D) y 2 +x 2 =D

5. Let’s try another one! Ex. dy/dx= ky 1. Move all x to one side and all y to one side: 1/ y dy= k dx 2. Antidifferentiate both sides: ∫ 1/ y dy= ∫ k dx In( y )= kx+C 3. Isolating y: y= e (kx+c) y= e kx x e c (e c is another constant, so let’s name it D) y= De kx

6. Finding specific antiderivatives using initial conditions By solving the Dif EQ using separation of Variables, we can find a general solution to the equation. But a family of similar curves can have the same derivative… So if we need to find a particular solution, we would need to find a more specific equation that contains a particular set of points

7. Find the solution of the Dif EQ that satisfies the given initial condition Ex. dy/dx= ky for which y(0)=3 and y(5)= 10 1. Move all x to one side and all y to one side: 1/ y dy= k dx 2. Antidifferentiate both sides: ∫ 1/ y dy= ∫ k dx In( y )= kx+C 3. Isolating y : y = e (kx+c) y = e kx x e c (e c is another constant, so let’s name it D) y = De kx is the general solution.

8. 4. Plug in y(0)= 3 to find D : 3= D e k(0) 3= D 5. Plug in y(5)= 10 to find k 10=3e k (5) 6. Solving for k: 10/3= e 5 k Ln(10/3)= 5 k 0.241~ k Particular solution: y= 3e 0.241x

9. Exponential Functions and y’=ky Example from book p. 428 #23 The number of bacteria in a certain culture increases at a rate proportional to the population. The initial number is 20000 and grows to 48000 in three hours.

10. Data interpretation: (0, 20000) (3, 48000) x= time in hours, y= number of bacteria So… at time= 0, the bacteria population was 20000 3 hours later, the population grew to 48000

11. A) find an equation that models the growth rate of bacteria. dy/dx= Growth rate y= population of bacteria So since “The number of bacteria in a certain culture increases at a rate proportional to the population…” dy/dx= ky

12. dy/dx= ky Use Separation of Variables 1. Move all x to one side and all y to one side: 1/ y dy = k dx 2. Antidifferentiate both sides: ∫ 1/ y dy = ∫ k dx In( y )= k x +C 3. Isolating y : y = e (kx+c) y = e kx x e c (e c is another constant, so let’s name it D) y = De k x

13. Plug in data points to solve for particular solution: y= De kx (0, 20000) (3, 48000) 1. Plug in (0, 20000) to solve for D 20000=De k(0) 20000=D 2. Plug in (3, 48000) to solve for k 48000=20000e k(3) 48000/20000= e k(3) Ln(2.4)/3= k 0.292~k

14. Therefore… Since D= 20000 and k= 0.292 Sub in variables: y=De kx y=20000e 0.292 X

15. By now you should know how to… Solve differential equations and use in modeling Find specific antiderivatives using initial conditions Study equation y’=ky and exponential growth