Question

Lokisan! Please help me with more differential equations! :)

Answer 1

22xdy+11ydx+3ydy=0

Answer 2

Loki! haha I'm back with more questions. :)

Answer 3

Hehe, I see...

Answer 4

Are you studying exact differential equations?

Answer 5

I don't think so. I know what the answer is and its not a number. Its the formula of the constant

Answer 6

Can you wait a little bit - I'm just finishing another post.

Answer 7

Sure

Answer 8

22xdy+11ydx+ydy=0 is the actual problem, Sorry

Answer 9

ok

Answer 10

i need paper

Answer 11

haha alright

Answer 12

oh, i don't have her question.

Answer 13

Scotty, this looks worse than it is (sorry, I have someone else I'm helping here too).

Answer 14

I'm hoping so. I think it fits into the trick of d(xy)=dy+ydx am I correct?

Answer 15

Hang on...I misread it...I have to work something out. Sorry, I'll just finish with this other person and I won't be distracted.

Answer 16

Ok so the problem is 22xdy+11ydx+3ydy=0

Answer 17

I wish that 22 wasn't there. You sure it's not 11?

Answer 18

Positive :/

Answer 19

I know what the answer if, if that helps.

Answer 20

yeah punch it in

Answer 21

Its \[11xy^2+y^3=C\]

Answer 22

Sorry, I keep being called away.

Answer 23

It's alright..

Answer 24

I'm thinking a substitution method. Just one sec...again called away...sorry. Won't be forever.

Answer 25

I think I may have your method...just let me check properly.

Answer 26

Ok cool

Answer 27

I keep being called away! I'll just let you know what I'm thinking and you can play with it as well.

Your equation, when rearranged, looks like this\[y'=-\frac{11y}{22x+y}=-\frac{11(\frac{y}{x})}{22+(\frac{y}{x})}\]after dividing the numerator and denominator by x.

Answer 28

This is in the form for a type of differential equation that can be solved when we know that y' is some function of y/x; that is,\[y'=F(\frac{y}{x})\]

Answer 29

We make the identification,\[v(x)=\frac{y}{x}\]so that\[y=xv\]

Answer 30

and we take it from there. Is this familiar?

Answer 31

Ok let me try to pmess with it

Answer 32

Ok instead of that one... can you help me solve y'-9y=8e^x

Answer 33

I'm doing this one for a timed quiz

Answer 34

I figured out your other problem - I made an arithmetic error, but the method is correct.

Answer 35

Yeah I'll do the second.

Answer 36

Ok I think I saw where you were going with it and I gutted my way through it. Thanks

Answer 37

Okay, integrating factor one - do you know?

Answer 38

Not sure what that means..

Answer 39

\[y'-9y=8e^x \rightarrow \frac{d (\mu y)}{dx}=\mu 8e^x\]

Answer 40

\[\mu = e^{\int\limits_{}{}-9dx}=e^{-9x}\]

Answer 41

so

Answer 42

im sorry that should be
y"

Answer 43

\[e^{-9x}y=e^{-9x}.8e^x+c \rightarrow y=ce^{9x}-e^x\]

Answer 44

y''-9y=8e^x ?

Answer 45

Yes, sorry for the confusion. I miss typed.

Answer 46

right
do you just want the answer for the quiz and then the method

Answer 47

Well. I want to learn but first and foremost I really need to pass this quiz

Answer 48

I am writing it all down to try to understand though.

Answer 49

\[y=c_1e^{3x}+c_2e^{-3x}-e^{x}\]

Answer 50

I'll explain...

Answer 51

Are you sure the + isn't a - between the first 2 terms?

Answer 52

You have a second order differential equation which is linear, inhomogeneous (because the RHS is a function of x only) with constant coefficients.

You need to solve inhom. equations by

1) solving the equivalent homogeneous equation first
2) finding a particular solution

Answer 53

+

Answer 54

In the end, it doesn't matter because it's absorbed into the constant.

Answer 55

The only option I have with a + sign in it is...
-2e^3x+e^-3x-e^x

Answer 56

Ok so e^3x=e^-3x-e^x is the same thing?

Answer 57

Oh, do you have initial conditions?

Answer 58

Sorry when x=0

Answer 59

Your equation has it's constants solve for

Answer 60

Oh...you need to give me the whole question ;)

Answer 61

Wait.... y=0 and y'=8 and x=0... I'm sorry I've been up all day doing this

Answer 62

It's second order, so there should be 2 conds.

Answer 63

ok

Answer 64

c_1 = 2 and c_2 = -1

Answer 65

\[y=2e^3x-e^{-3x}-e^x\]

Answer 66

Yep that works!

Answer 67

Good

Answer 68

Thanks so much... I really feel low just asking for answers but I plan on taking the class again. If I fail I have to pay the whole tuition and I've been studying so hard and just not getting it lately :(

Answer 69

It's because you're having to do it on your own - you're not being shown the ropes. Don't punish yourself.

Answer 70

trying not to. Could you help me with 1 or 2 more? Then I'll leave you alone.

Answer 71

yes

Answer 72

How long do you have to do the quiz?

Answer 73

i panic in tests - you shouldn't have told me, lol

Answer 74

4xydx=(x^2+4)dy

and that is the whole problem and not leaving anything out :)

Answer 75

It's separable.

Answer 76

\[4xydx=(x^2+4)dy \rightarrow \frac{4xy}{x^2+4}=\frac{dy}{dx} \rightarrow \frac{4x}{x^2+4}dx=\frac{dy}{y}\]

Answer 77

You can integrate from there.

Answer 78

Ok let me try

Answer 79

\[y=c(x^2+4)^2\]

Answer 80

yep thats what I got

Answer 81

Aesome

Answer 82

or, awesome

Answer 83

Ok I think this should be the last one... I think I got the others actually.

Answer 84

ok

Answer 85

dy+2ydx=e^(-4x)dx

Answer 86

It's an integrating factor one again.

Answer 87

Yeah those are really difficult for me. Seperatable ones are the ones I'm more comfortable with...

Answer 88

\[y=-\frac{1}{2}e^{-4x}+ce^{-2x}\]

Answer 89

I'm just skipping to the solution because of the test...

Answer 90

The closest answer I have that looks like that is e^(4x)+Ce^(2x)

Answer 91

hmmm...that's the solution to the equation you gave me...

Answer 92

let me write it in the formula so it looks better maybe it looks funny on text

Answer 93

\[dy+2ydx=e^{-4x}dx \rightarrow y'+2y=e^{-4x}\]?

Answer 94

\[dy-2ydx=6e^(5x) dx\]

Answer 95

that dan
5x is super script

Answer 96

so you gave me the wrong question before?

Answer 97

Im a failure sorry I'm rushing

Answer 98

\[y(x) = c_1 e^{2 x}+2 e^{5 x}\]