Question

Initial value problem

Answer 1

\[y' + y \tan (x) = \cos^2 (x) , y(\frac{ \pi }{ 4 }) = \frac{ 1 }{ 2}\]

Answer 2

\[y' = \cos^2(x) - ytan(x)\]

Answer 3

\[\int\limits_{}^{} \cos^2(x) - y \tan(x) dx\]

Answer 4

is this how I should solve this problem? This is new to me

Answer 5

thats not how i remember this going

Answer 6

youve got a y stuck in there, you cant just play it like that

Answer 7

Yeah I noticed it. I'm unsure as to how to proceed though.

Answer 8

let y = e^(rx), what is y'?

Answer 9

or ive seem some:

let y=x^n, and y' is?

Answer 10

n x^(n-1)

Answer 11

sub in, not sure which one is more apt for this tho

Answer 12

solve the homogenous part first

y' + ytan(x) = 0

which may be seperable

Answer 13

\[y'+ytan(x)=0\]
\[\frac{y'}{y}=\frac{-sin(x)}{cos(x)}\]
does this look doable?

Answer 14

y'/y=-tan(x)?

Answer 15

yep

Answer 16

but might want to rewrite the tan, its more noticable that way

Answer 17

alright then I would end with y'/y = -(sin(x)/cos(x)) + cos^2(x)

Answer 18

no, work the homogenous (the =0) first to get a y_h solution

Answer 19

\[y'+yf(x)=0\]
\[\int y'/y=-\int f(x)\]
\[ln(y)=F(x)+C\]
\[y=Ce^{F(x)}\]

Answer 20

ask if somethings confusing

Answer 21

from ln(y) = F(x) +C how does it changes to y = Ce^F(x)?

Answer 22

base e both sides and recall
\[a^{i+j}=a^i~a^j\]

Answer 23

as such e^C is just some arbitrary constant

Answer 24

if C was there then it would just have been e^F(x) right?

Answer 25

if C was not there, then yes

Answer 26

ok, got it.

Answer 27

so, how do we apply this to your problem?

Answer 28

shouldn't it be -F(x) + C?

Answer 29

thats not really important, it was a generality

Answer 30

what is the integration of -sin(x)/cos(x) ?

Answer 31

ln|cos(x)|

Answer 32

good so we have

ln(y) = ln(cos(x)) + C

e^(ln(y)) = Ce^(ln(c0s(x)))

y = ________

Answer 33

y=Ce^(ln|cos(x)|)

Answer 34

and e^ln(this) = this

Answer 35

shouldn't it be y = Ce^(ln|cos(x)|)?

Answer 36

which makes y = C*cos(x)

Answer 37

yes thats fine

now that we have y_h, the homogenous solution of y, we want to find the particular solution y_p some texts call this y_c

Answer 38

I don't have knowledge as to what is "y_h" and so on.

Answer 39

y_1 and y_2 maybe?

Answer 40

nope, never saw that

Answer 41

oh well, then this is something new :) or maybe there was another way to approach it ... but i dont see how

we can proceed if you like

Answer 42

sure

Answer 43

now youll learn some day that that the solution y, is a linear combination of y_1 and y_2, or whatever they want to start you off with: later they call it a homogenous and a particular solution (y_h and y_p) such that:

y = ay_h + by_p is the full solution

Answer 44

homogenous just means that the y's setup is equal to 0 and solved

youll learn that y_p is a multiple of y_h by some function u(x) or whatever name youwant to call it; ill call it A(x) and write

y_p = A cos(x) , realizing that A is a function of x not a constant

y'_p = A' cos(x) - A sin(x)

and we sub it into the equation and solve

Answer 45

\[(y_p)'+(y_p)tan(x)=cos^2(x)\]
\[A'cos(x)-Asin(x)+Acos(x)tan(x)=cos^2(x)\]
\[A'cos(x)-\underbrace{\cancel{Asin(x)+Asin(x)}^{0}}_{homogenous~part}=cos^2(x)\]
\[A'cos(x)=cos^2(x)\]
can you solve A' ?

Answer 46

this is basically A' = cos(x) but what happened to the past answer y= C*cos(x)?

Answer 47

its still a part of y .. we are solving a particular part of y now y_p

Answer 48

y = C cos(x) + y_p

y = C cos(x) + A(x) cos(x)

etc ...

Answer 49

you see that part i crossed out, the homogenous part? that is taken care of as we solve for y_p .... we expect it to go to zero in the solving process since it equalszero when we found it the first time

Answer 50

I see but where does the A(x) comes from?

Answer 51

its a function that multiples to y_h to make it an independant vector, but thats something that youll learn or forgot to learn

it makes the particular solution valid so that its not just some constant multiple of y_h

Answer 52

is quite strange because the professor didn't go over this although I missed one day but there is no way she taught this in an hour lol

Answer 53

watch

A' = cos(x)

A = sin(x) + C

y = y_h + y_p

y = C cos(x) + (sin(x)+C) cos(x)

y = C cos(x) + Ccos(x) + sin(x)cos(x)

y = 2C cos(x) + sin(x)cos(x)

but 2C is just an arbitrary constant ... C eats it up


y = C cos(x) + sin(x)cos(x)

Answer 54

now we need to find the C is when we have the initial condition :)

Answer 55

hmm I see what's going on. is y_h as y underline h?

Answer 56

or is there a symbol or something to it

Answer 57

\(y_h\)
h for homogenous
p for particular
or c for complment

and before this part its just 1 and 2

Answer 58

i dont know how well another idea i have would work, its called an integrating function; pretty much the A(x) they call u(x) and work a reverse product rule

Answer 59

by any chance does this fall under Sequences and series - infinite sequences of real numbers . . .

Answer 60

\[\mu(x)[y'+ytan(x)=cos^2(x)]\]
\[\mu(x)y'+\underbrace{tan(x)\mu(x)}_{=\mu'(x)}y=\mu(x)cos^2(x)\]
....
i wouldnt know what it falls under :) ive been out of college too long to remember specific chapters and such

Answer 61

is it \[\int\limits_{}^{}\frac{ y' }{ y } really = \ln(y)? \int\limits_{}^{}\frac{ \frac{ dy }{ dx} }{ y }\]

Answer 62

if u' = u tan(x)

u'/u = tan(x)

ln|u| = ln|-sin(x)| , we dont need all functions so when C=0 its fine

u = -sin(x) works

\[\mu(x)y'+\underbrace{tan(x)\mu(x)}_{=\mu'(x)}y=\mu(x)cos^2(x)\]
\[\mu(x)y=\int\mu(x)cos^2(x)~dx\]
\[y=\frac{1}{\mu(x)}\int\mu(x)cos^2(x)~dx\]
\[y=-\frac{1}{sin(x)}\int\sin(x)cos^2(x)~dx\]

Answer 63

forgot a negative, drop the one there lol

Answer 64

ln(y) , chain rule, derives to y'/y

Answer 65

right?

Answer 66

derivative of ln(y) isn't 1/y?

Answer 67

no, y wrt.x

not y wrt.y

Answer 68

\[\frac{ 1 }{ y} dy\]

Answer 69

if no 'wrt.' is stated then its implicitly considered a function of x or time or something else.

the derivative of y is y'
the derivative of ln(x) is x'/x
the derivative of cos(u) is -u' sin(u)

etc ...

Answer 70

oh ok ok

Answer 71

\[\frac{dy/d*}{y}\]

Answer 72

if we know what it is 'with respect to' then we can determine if the (') variable is equal to 1 or not :)

Answer 73

wrt.x: x' = dx/dx = 1

wrt.x: y' = dy/dx = whatever it gets to be

wrt.t: u' = du/dt = whatever it gets to be

Answer 74

going back to : y_p = A cos(x) , realizing that A is a function of x not a constant

y'_p = A' cos(x) - A sin(x) how do you determine which is y_p and y'_p?

Answer 75

we solve for y_h by setting cos^2(x) tp 0 in this case

then setting y_p = A y_h w can determine y_p for the particular solution of cos^2(x)

Answer 76

the solution to y_p will contain y_h in it once its solved, which i demonstarted by pulling it out and combining it to the y_h we already had there

Answer 77

ok so first was the part where we solve y' + ytan(x) = 0

Answer 78

which ended with the result of y = Ccos(x)

Answer 79

notice that when A is equal to some constant, then y_p = C y_h is already a solution

Answer 80

not y, y_h

Answer 81

\[(y_h)'+(y_h)tan(x)=0\]
from this we determined y_h = C cos(x), a constant multiple of cosine

Answer 82

ohhh ok that makes a bit more sense

Answer 83

once yh is solved then comes yp?

Answer 84

yes, once we know one solution to it, the homogenous solution, then we can use that to determine a particular solution as a multiple of y_h and some function of x

Answer 85

correct

Answer 86

refresh to get rid of the diamonds .. its a copy paste decoding error for some reason

Answer 87

ok so we have y_h = Ccos(x) and A' = cos(x)

Answer 88

right, plugging in y_p and y'_p we determined A' = cos(x)

Answer 89

how is y_h and A' related

Answer 90

A' = cos(x)

A = sin(x) + C

y = y_h + y_p

Answer 91

nvm I see it

Answer 92

notice:

y_p = (sin(x)+C)cos(x), we would expect it to contain y_h since its made from it

Answer 93

y_h = Ccos(x) and y_p = sin(x)+C

Answer 94

where does the other cos(x) after (sin(x)+C) come from?

Answer 95

from the build we did

y_h = C cos(x)

let y_p be some multiple of y_h ... instead of a C, make it a function

y_p = A(x) cos(x) , now this is equal to y_h when A(x) is a constant, right?

Answer 96

yes

Answer 97

when we solve for y_p, it will inherently have y_h built into it

y_p = A cos(x) ; A = (sin(x)+C)
^^^
theres it is

y_p = sin(x)cos(x) + Ccos(x)

which is really just the solution we wanted all along

Answer 98

C is not = to Ccos(x) though