Question

For first order linear differential equation problems, where do I start? How do i use P(x) and Q(x) and u(x)???

Answer 1

maybe start wid an example problem :)

Answer 2

I have a test wednesday morn..... still not understanding! and yes, I know about the IF method but not quite absorbing

Answer 3

ok uhm (3x+sin2x)dx -dy=0

Answer 4

And it just says to solve ^^

Answer 5

your goal is to find \(y\)

Answer 6

when it just says "solve"

Answer 7

\( (3x+\sin2x)dx -dy=0 \)

\( (3x+\sin2x)dx = dy\)

Integrate both sides

\( \int (3x+\sin2x)dx = \int dy\)

\( \int (3x+\sin2x)dx = y\)

Answer 8

we're done. we have "solved" y

Answer 9

I know how to solve that way, but... is there a way to solve it with all the fancy P(x) and Q(x) and u(x) stuff? I think this may be slightly connected to the IF thing that we talked about yesterday before your lunch ^^

Answer 10

sure u can solve it by finding IF and multiplying it both sides


but nobody does that cuz this can be easily solvable by separating variables

Answer 11

if u want to do it, we can do it... just to give u practice with IF method ;)

Answer 12

step1 : put the given equation into "standard linear differential equation form" :

\(\large y' + p(x) y = q(x)\)

Answer 13

hmmm what do I do if I come across something like dy/dx=3y+sin2x??? there's two terms, and I can't really transfer a term over because that would mean excluding it from the dx or dy that it's supposed to be with...

Answer 14

this is a new equation eh ?

Answer 15

Are we done with the first equation ?

Answer 16

OoO no I just rearranged it in a failed attempt to make it into the linear form

Answer 17

nope, u changed the second equation.
look closer

Answer 18

your original equation : (3x+sin2x)dx -dy=0
your rearranged equation : dy/dx=3y+sin2x

Answer 19

they dont match
(you have replaced 3x with 3y)

Answer 20

which one is correct ?

Answer 21

oh whoops!!! It's 3y~

Answer 22

cool :) then none of the discussion so far is meaningless

Answer 23

you cannot solve it by using "variable separation method"

Answer 24

you will hve to use the IF method

Answer 25

XP srry haha yeah~

Answer 26

lets start from here :
dy/dx=3y+sin2x

Answer 27

oh hey! haha accidental success!

Answer 28

you wanto change it to form :
\(y' + p(x) y = q(x) \)

Answer 29

:) replace dy/dx wid y' first

Answer 30

so then just take the 3y and slap it over onto the left side?

Answer 31

\( \large \frac{dy}{dx}=3y+\sin2x \)

\( \large y'=3y+\sin2x \)

\( \large y'+ (-3)y = \sin2x \)

Answer 32

yes ! compare it wid the standard form :
\(y' + p(x) y = q(x)\)

Answer 33

p(x) = ?
q(x) = ?

Answer 34

p(x) is -3, and q(x) is sin2x~

Answer 35

yup !

step1 : find the IF

Answer 36

step2 : multiply IF both sides and compress the left hand side

Answer 37

step3 : integrate both sides

Answer 38

for step1, use the formula :

IF = \(\large e^{\int p dx}\)

Answer 39

ohh ok

Answer 40

for our present problem,

IF = \(e^{\int -3 dx} = e^{-3x}\)

Answer 41

you're done with step1

Answer 42

next, multiply the IF both sides of ur rearranged equation in standard form

Answer 43

okay, so I got to the part where I've already condensed the left side and put the whole thing under an integration thing, and so now it looks like this: ye^-3x=(integral sign) e^-3x (sin2x) -----is there supposed to be a dx here???

Answer 44

\(\large y'+ (-3)y = \sin2x\)

multiply IF = \(e^{-3x}\) both sides


\(\large e^{-3x}y'+ e^{-3x}(-3)y = e^{-3x}\sin2x\)

\(\large (ye^{-3x})' = e^{-3x}\sin2x\)

Integrate both sides

\(\large \int (ye^{-3x})'~ dx = \int e^{-3x}\sin2x~dx\)

Answer 45

so the dx just kind of magically appears along with the integration signs?

Answer 46

And I have no idea how to integrate the right side...

Answer 47

\(\int dx\)


we're integrating both sides w.r.t x, so dx also comes

Answer 48

integrating right hand side is a different thing,
but, u are comfortable wid the mechanics of IF method right ?

Answer 49

so what about the left side? isn't it integration w.r.t. y?

Answer 50

nope, both sides we're integrating w.r.to x only

Answer 51

\(\large y'+ (-3)y = \sin2x\)

multiply IF = \(e^{-3x}\) both sides


\(\large e^{-3x}y'+ e^{-3x}(-3)y = e^{-3x}\sin2x\)

\(\large (ye^{-3x})' = e^{-3x}\sin2x\)

Integrate both sides

\(\large \int (ye^{-3x})'~ dx = \int e^{-3x}\sin2x~dx\)


\(\large ye^{-3x} = \int e^{-3x}\sin2x~dx\)

Answer 52

ohhh ok, so it's much like putting a derivative onto the equation~

Answer 53

yes, lets look at integrating right hand side

Answer 54

u familiar wid "integration by parts" eh ?

Answer 55

wait so would the derivative sign on the left side come off even w.r.t y?

Answer 56

we're interested in finding "y" as a function of x,
so "integrating with respect to x" annihilates the "derivative with respect to x"

Answer 57

you always integrate with respect to the independent variable : x or t

Answer 58

hmmm so then the left side isn't mulitplied by dy/dx?

Answer 59

\(\large \int (ye^{-3x})'~ dx = \int e^{-3x}\sin2x~dx\)

Answer 60

let me ask u a question :)

wat does \(\large (ye^{-3x})'\) represent ?

Answer 61

and so the left side is [ ye^-3e ]dy/dx? or no???

Answer 62

\(\large (ye^{-3x})'\) is same as \(\large \frac{d}{dx}(ye^{-3x})\)

Answer 63

just a notation thing,
for me the first notation looks neat

Answer 64

so i use that,
if u prefer the d/dx 's u may use that as well

Answer 65

oh hmmm haha ^^ I thought the first thing meant dy/dx, so I was confused~

Answer 66

ok so then the left side is left bare with nthn, and now on to the right side?

Answer 67

\(\large y'+ (-3)y = \sin2x\)

multiply IF = \(e^{-3x}\) both sides


\(\large e^{-3x}y'+ e^{-3x}(-3)y = e^{-3x}\sin2x\)

\(\large (ye^{-3x})' = e^{-3x}\sin2x\)

Integrate both sides

\(\large \int (ye^{-3x})'~ dx = \int e^{-3x}\sin2x~dx\)

\(\large ye^{-3x} = \int e^{-3x}\sin2x~dx \)

Answer 68

you are at peace wid everything so far eh ?

Answer 69

yesh, more or less so ^^

Answer 70

the right side is just a "integration problem" :

you can do it by parts very easily :
http://www.wolframalpha.com/input/?i=%5Cint+e%5E%28-3x%29+sin%282x%29+dx

Answer 71

WOAH COOL wat is thissss

Answer 72

\(\large y'+ (-3)y = \sin2x\)

multiply IF = \(e^{-3x}\) both sides


\(\large e^{-3x}y'+ e^{-3x}(-3)y = e^{-3x}\sin2x\)

\(\large (ye^{-3x})' = e^{-3x}\sin2x\)

Integrate both sides

\(\large \int (ye^{-3x})'~ dx = \int e^{-3x}\sin2x~dx\)

\(\large ye^{-3x} = \int e^{-3x}\sin2x~dx \)


\(\large ye^{-3x} = \frac{-1}{13}e^{-3x}[3\sin(2x) + 2\cos(2x)] + c \)

Answer 73

we're done

Answer 74

QQ I still don't understand how wolf ram got the answer tho...

Answer 75

dont wry, we're going to work it now

Answer 76

\(\int e^{-3x} \sin (2x) ~dx\)

Answer 77

familiar with "by parts" ?

Answer 78

kind of~ but I only know the + ones~ where you split the terms?

Answer 79

\(\int uv' = uv - \int u'v \)

Answer 80

does that look familiar /

Answer 81

?

Answer 82

can't say it does... |:

Answer 83

maybe it's a BC topic...

Answer 84

^^ well, thank you so much for your help~! I'll be going to bed now! it's 4am already... *yawn* bibi^^ have a good... lunchtime? LOL I meant afterenon ^^

Answer 85

\(\large I = \int e^{-3x} \sin (2x) ~dx \)

By parts : \(u = \sin(2x), v' = e^{-3x}\) , gives


\(\large I = \sin (2x) \frac{-e^{-3x} }{3} - \int 2\cos (2x) \frac{-e^{-3x} }{3} ~dx \)


\(\large I = \sin (2x) \frac{-e^{-3x} }{3} +\frac{2}{3} \int \cos (2x) e^{-3x} ~dx \)

Answer 86

do it by parts again for that integral again

Answer 87

\(\large I = \sin (2x) \frac{-e^{-3x} }{3} +\frac{2}{3} \int \cos (2x) e^{-3x} ~dx \)

\(\large I = \sin (2x) \frac{-e^{-3x} }{3} +\frac{2}{3} \left( \cos(2x) \frac{-e^{-3x} }{3} - \int (-2\sin (2x))\frac{-e^{-3x} }{3} \right) \)


\(\large I = \sin (2x) \frac{-e^{-3x} }{3} -\frac{2}{9} \cos(2x)e^{-3x} - \frac{2}{3} \int \sin (2x)e^{-3x} \)


\(\large I = \sin (2x) \frac{-e^{-3x} }{3} -\frac{2}{9} \cos(2x)e^{-3x} - \frac{2}
{3} I \)

Answer 88

solve \(I\)

Answer 89

have good sleep :)

Answer 90

^^Thats one very long method :/

Answer 91

Another easy/super fast way(my fav) is to work out \(\int e^{-3x}\sin(2x)dx\) using complex numbers :


\(\large \mathbb{\int e^{-3x} \sin(2x) dx = Img \left( \int e^{-3x} e^{i (2x)}dx\right) }\)

\(\large \mathbb{~~~~~~~ = Img \left( \int e^{x(2i-3)} dx\right)}\)

\(\large \mathbb{~~~~~~~ = Img \left( \frac{e^{x(2i-3)}}{2i-3} \right) }\)

\(\large \mathbb{~~~~~~~ = Img \left( e^{-3x}\frac{2i+3}{13}[ e^{i(2x)}] \right) }\)

\(\large \mathbb{~~~~~~~ = Img \left( e^{-3x}\frac{2i+3}{13}[ \cos(2x) + i\sin(2x)] \right) }\)

\(\large \mathbb{~~~~~~~ = \frac{e^{-3x}}{13} [ 2\cos(2x) + 3\sin(2x)] }\)