Question

Solve the homogeneous differential equation 2x dx + (y-3x)dy = 0

Answer 1

@Loser66

Answer 2

Have you done these yet?
Bạn đã làm những chưa?

Answer 3

roi

Answer 4

hahahaha.... interesting. !!!

Answer 5

does that mean no? lol i don't speak vietnamese i just have a translator

Answer 6

yes! I did

Answer 7

how can you put it there?

Answer 8

copy and paste from translate.google.com

Answer 9

you are...but interesting

Answer 10

so how do you solve @Loser66 there's two approaches to homogenous de's and how do you know if it is homogeneous?
vậy làm thế nào để bạn giải quyết @ Loser66 có hai phương pháp tiếp cận đồng nhất de và làm thế nào để bạn biết nếu nó là đồng nhất?

Answer 11

hahahaah... I know how to consider whether it's exact or not, friend. there are 2 ways to solve.separate or substitute y =Vx

Answer 12

am I right?

Answer 13

yes you substitute y=vx... the way to tell is a bit hard to explain, but i'll try

Answer 14

so say the equation is of the form M(x,y)dx + N(x,y)dy
If you rewrite the equation as M(xt,yt)dx + N(xt,yt)dy and the same factor of t factors out of the equation, then its homogenous... does that make sense?

Answer 15

So M(xt,yt) = 2xt = t(2x)
and N(xt,yt) = yt -3xt = t(y-3t)
both factor out a t so its homogenous...

Answer 16

I have a way to figure out the solution: let see.

Answer 17

2xy -3x^2 =C, it's mine

Answer 18

I got something way different

Answer 19

ok, that's the trick. let ,me redo, heheheh

Answer 20

i didn't check if that works, (i hate the calculus of checking it in this form heh)

Answer 21

2x dx +(y - 3x) dy = 0 let y = ux . So dy = u dx + x du and use substitution.
2x dx + (ux - 3x) ( u dx + x du) = 0
2 dx + (u - 3)(u dx + x du) = 0
2 dx + u^2 dx + ux du - 3u dx - 3x du = 0
(u^2 -3u +2)dx = ((3x -ux) du
(u-2)((u-1)dx = x (3 - u) du
now separate the variables.
dx/x = (3-u) / (u-2) (u -1) du
the left hand side is ln x + C. The right hand side will require partial fractions.
It will be of the form A/ (u-2) + B/ (u-1).
Solving for A and B you will get A = 1 and B = -2 .
So integrating the right hand side means integrating 1/(u-2) + -2/(u-1). We obtain ln(u-2) -2ln(u-1).
We now have:
lnx + C = ln(u-2) -2ln(u-1)
Substitute back in that u = y/x (from the beginning substitution)
We now have:
lnx + C = ln((y-2x)/x) -2ln((y-x)/x)
Apply log properties:
lnx + C = ln(y-2x) - ln x -2ln(y-x) + 2ln x
C = ln(y-2x) -2ln(y-x)

Answer 22

\[\frac{ y^2-2xy+x^2 }{ y-2x }=C\]
is the answer but i was looking to see if you did it differently @loser66

Answer 23

@rrr same answer as mine, except you need to raise both sides to e to cancel the logs

Answer 24

our answers are reciprocals of each other though, hrmm

Answer 25

to me, I check whether it is exact or not. yours is not.
then find \[\mu\] to make it exact. by formula. base on that differential equation which has exact solution find int respect to y of M and int respect to x of N. pick like term add them together to get the solution.

Answer 26

It is OK that they are reciprocals since they would equal an arbitrary constant. The constants that they equal are not the same constant. They are also reciprocals.

Answer 27

Let me make sure I understand your question. You want to know what on your particular solution, you have a t in one solution and not in the other. when you are making it look like your RHS of the nonhomogeneos DE.

Answer 28

this is another one

Answer 29

When you find your from of particular solution when you working with the method of undetermined coefficient, your form is determined not only by the RHS of your DE but also by the complementary solution of the DE.

Answer 30

oh yes, I got what you mean, because the solution from the LHS has the form of e^-t already, so, the nonhomogeneous part must be At^-t , right?
On example 1, the LHS doesn't have the form of e^2t so no need to have t at the RHS, right?

Answer 31

Yes, you have it.

Answer 32

the last one, please, please, please!!

Answer 33

Gotta go put my kids to bed. Sorry.

Answer 34

what's the question?

Answer 35

at my attachment above

Answer 36

\[y"-3y'-4y=-8e^tcos(t)\] this?

Answer 37

yes.

Answer 38

so we need the particular solution

Answer 39

cos (2t)

Answer 40

\[\huge y"-3y'-4y=-8e^tcos(2t)\]

Answer 41

ok got it, give me a sec ok?

Answer 42

sure. I am waiting

Answer 43

been just a bit since i've done nonhomogenous 2nd order equations, and there's a bunch of different ways to do this

Answer 44

I need just one. give me it, please,

Answer 45

\[y_p=e^t(Acos(2t)+Bcos(2t)\]
is what you used yes?

Answer 46

lol except the second is sin(2t)

Answer 47

I copied from my prof

Answer 48

yes, you are wrong there. but no big deal

Answer 49

ok so you know that the right hand side is going to be of that form

Answer 50

eliminate one constant in the front of e^t , right?

Answer 51

then?

Answer 52

yes we need a constant infront to... ok now we need to find the first and second derivatives and plug this all in to the ODE

Answer 53

which is not easy, can make mistakes

Answer 54

\[e^t(Acos2t+Bsin2t)=De^{(1+2i)t}\] right? from Euler's formula for complex number, right?

Answer 55

yes you can do that too

Answer 56

Oooookk, I got it now, thanks a ton .

Answer 57

ok :)

Answer 58

hey, you did something on justin.tv for fun?

Answer 59

Did i see something for fun? there's all kinds of things on justin.tv

Answer 60

oh no my channel on justin.tv isn't just for fun, its part of my business

Answer 61

what are you doing there?

Answer 62

hang on a second, i'm helping someone else