By Holum Kwok

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

So Assuming you know how to integrate a function…

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

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

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

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)= 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.

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

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 and grows to in three hours.

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

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

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

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

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

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