MAT 21B – Lecture 11 – Separable Differential Equations PDF

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MAT 21B – Lecture 11 – Separable Differential Equations – 01/30/2019   

Differential Equation – an equation where both a function and its derivative appear. The goal is to solve for the function. If we have enough initial conditions, there is a single solution. If not, we get an infinite family of solutions. Example: Solve for y when

dy =5 y . dx

First rearrange the differential equation so that





is equivalent to

1 dy =5 . Take the integral of both sides of the equation, resulting in y dx 1 dy ∫ y dx dx=∫ 5 dx . Here the differential, dx cancel on the left side of the 1 equation leaving us with ∫ dy=∫ 5 dx . Integrating both sides, we have y ln|y |=5 x +C . A single constant takes care of both constants from the integration. Solving for y, e ln |y |=e 5 x+C and with further manipulation, |y|=e c e 5 x or y =± e c e 5 x . Let A=±e c so that our final answer is 5x 5x is an infinite family of solutions to the differential y= A e . So y= A e dy dy d 5x 5x =5 y . Check: = ( A e ) = A e ∗5=5 y . equation dx dx dx dy =5 y with As a continuation of the example above, find the solution to dx y ( 0)=17 . Applying y (x)= A e5 x , we have 17= y ( 0 )= A e 5∗0= A and so y (x)=17 e5 x



dy =5 y dx

is the unique solution, since we had enough initial conditions.

dy =f (x , y) , the method of dx integrating both sides works as long as we can write f ( x , y ) =g ( x ) h( y ) in which g ( x ) is a function of x only and h( y ) is a function of y only. This is true because we can always divide both sides by h( y ) to get 1 dy =g (x) . Then integrate both sides with respect to x such that h( y ) dx In general, if we have the differential equation,

1

dy

∫ h( y ) dx dx=∫ g ( x) dx

to get

1

∫ h( y ) dy =∫ g (x)dx

and then find the

two integrals separately. 

Definition: The differential equation

dy =f (x , y) is separable if dx

f ( x , y ) =g ( x ) h( y ) 

Example 2: Solve for

y (x)

if

2 y ' ( x ) + y ( x ) cos ( 2 x )=0 .

This can be rewritten as

dy 2 + y cos ( 2 x ) =0 , which is equivalent to dx

dy =− y 2 cos (2 x ) . Choose g ( x )=−cos(2 x) dx that

1

∫ y2 dy=∫−cos(2 x )dx

and

h ( y )= y 2 . It follows

. Integrating both sides, we have

−1 −sin(2 x) = + C and multiplying both sides of the equation by -1, then 2 y 1 sin (2 x ) −2C 1 sin(2 x ) and = −C . To solve for y, first rewrite this as = 2 y 2 y 2 2 or y= . Note that in the previous example, sin ( 2 x) −2C sin ( 2 x ) + A g ( x )=5 and h ( y )= y .

so





y=

What is the relationship between exponential change and separable differential equations? In exponential growth/decay, the increase/decrease of a quantity depends on the quantity itself. Example: Suppose p ( t ) is the world population at time t. Then p' (t ) represents the population growth rate at time t. So

p ( t)

on each other, which results an equation involving both 

and

p' (t )

p( t ) and

depend

p' (t ) .

Example 2: Suppose b(x ) is the money in your bank account at time x. Then b ' ( x ) represents the rate at which you collect interest at time x. So b(x )

b ' ( x ) depend on each other, which results an equation involving both b(x ) and b ' ( x ) .

and 

Section 7.4 – Problem 26: The processing of raw sugar has a step called “inversion” that changes the sugar’s molecular structure. Once the process has begun, the rate of change of the raw sugar is proportional to the amount of raw sugar remaining. If 1000 kg of raw sugar reduces to 800 kg of raw sugar during the first 10 hours, then how much raw sugar will remain after 14 hours? Let s (x) represent the amount of sugar at time, x in kg. We are given the initial conditions

s ( 0 )=1000

and

s ( 10 )=800 . This rate of change is

ds =ks ( x )=ks . Let h ( s )= s and g ( x )=k . Applying the dx 1 ds dx= g ( x ) dx 1 general rule, ∫ , we have ∫ ds=∫ k dx and ∫ s h(s) d x kx integrating both sides, ln |s|=kx +C to get s= A e . Next, we need to find the values of A and k . This can be done by first using one of the initial value conditions, s ( 0)=1000=A e k∗0 = A . Therefore, s ( x ) =1000 ekx . To find represented by

the value of k, use the second initial value condition such that

s ( 10 )=1000 e 10 k =800 . Rearranging it, we have

e 10k =

4 . Take the 5

logarithm of both sides such that with

k=

ln e 10 k =ln (

4 4 ) resulting in 10 k=ln( ) 5 5

()

4 1 =0.1 ln(0.8) . Thus, s (x)=1000 e0.1 ln(0.8 )x . Finally, the ln 5 10

amount of raw sugar that will remain after 14 hours is given by (

) ∗14

s ( 14 )=1000 e 0.1 ln 0.8

≈ 732 kg....