Question

Differential Equation PLEASE HELP!

Answer 1

\[\frac{ (y-xy') }{ (x+yy' }\]

Answer 2

=2

Answer 3

\[\frac{ (y-xy') }{ x+yy') }=2\]

Answer 4

That's not ODE, is it?

Answer 5

@rvc?

Answer 6

@IrishBoy123 ?

Answer 7

rearrange it so that it reads y' = ...whatever you get

then we'll call the result y' = f(x, y)

if then f(tx, ty ) = f(x,y), then it is a homo and you are good to go.

Answer 8

I got \[y'=\frac{ (2x+y) }{ (y-2x) }\]

Answer 9

you mean to write y' as f(2x,y)?

Answer 10

is that not upside down?!

or have i misread the original equation.

Answer 11

Was I wrong even with that?

Answer 12

\[\frac{ y - xy' }{ x + y y' } = 2\]

y - xy' = 2(x + yy')
y - xy ' = 2x + 2yy'
y'(2y +x) = y - 2x

Answer 13

ok, but what m I doing from here?

Answer 14

am

Answer 15

y' = f(x,y) = (y - 2x)/(2y +x)

yes?

Answer 16

so the homo test is:

does f(kx,ky) = f(x,y)

for any number k. can you establish that that is true, because it means we can nail it using a homo sub.

Answer 17

so k is 2 right?

Answer 18

no, k is any number. replace x by kx and y by ky in f(x,y) to get f(kx, ky). and find out if they are still equal.

Answer 19

yes

Answer 20

then homo sub right?

Answer 21

yes

Answer 22

so the fraction should be replaced with U ?

Answer 23

is that what you mean?

Answer 24

no, now you make the sub y = v x, ie replace all y's with vx.

then, because y = v x, you know that y' = xv' + v , and so you can make that sub too.

that is now separable in v and x, though there is some algebra to be done, moving the v's to the LHS and the x to the RHS.

bit messy but totally do-able..

Answer 25

I hope so, I'm doing it right now

Answer 26

@IrishBoy123 it's not working...

Answer 27

@phi?

Answer 28

once you get to
\[ y' =\frac{ (y - 2x)}{(2y +x)} \]
you do what you did on previous problems
y = ux
y' = u + x u'
how far do you get ?

Answer 29

yes

Answer 30

but it didn't work from there and I don't know why...

Answer 31

how far did you get ?

Answer 32

a mess, one stage after that

Answer 33

I'll try again one sec

Answer 34

\[u'x+u=\frac{ (ux-2x) }{ (2ux+x) }\]

Answer 35

you can factor out x from top and bottom on the right, and cancel.
write u' as du/dx and separate variables.
what do you get ?

Answer 36

\[\frac{ du }{ x }+u=\frac{ (u-2x) }{(2u+1)}\]

Answer 37

that does not look correct

Answer 38

I did common denominator, but it didn't help much tbh

Answer 39

first, just simplify the right side:
\[ \frac{ (ux-2x) }{ (2ux+x) } \]

Answer 40

ups

Answer 41

?

Answer 42

I got this \[\frac{ (2u+1) }{ (u-2) }du=\frac{ dx }{ x }\]

Answer 43

it seems better now ehe?

Answer 44

closer. But you have to be a lot more careful with your algebra
\[ u'x+u=\frac{ (ux-2x) }{ (2ux+x) } \\
\frac{du}{dx} x+u=\frac{ (u-2) }{ (2u+1) } \\
\frac{du}{dx} x=\frac{ (u-2) }{ (2u+1) } -u
\]

Answer 45

use a common denominator of 2u+1 on the right side

Answer 46

so what from here?

Answer 47

can you combine terms on the right side ?

Answer 48

yes \[\frac{ du }{ dx}=\frac{ -2-2u^2 }{ 2u+1}\]

Answer 49

ok, but remember there is a factor of x on the left side.

Answer 50

\[\frac{ (2u+1) }{ -2-2u^2 }du=dx\]

Answer 51

you started with
\[ \frac{du}{dx} x=\frac{ (u-2) }{ (2u+1) } -u \]
You lost the "x" ??

Answer 52

it's such a mess

Answer 53

messy or not, you need to keep the x

Answer 54

Should I start from the beginning?

Answer 55

This is ok
\[ \frac{ du }{ dx}x=\frac{ -2-2u^2 }{ 2u+1} \]
I would "flip" both sides
\[ \frac{ dx }{ x}\frac{1}{du}=\frac{ 2u+1} { -2-2u^2} \]

Answer 56

what for?

Answer 57

because it (should be) clear that we want dx/ x on one side
(surely not x/dx , right ?)
and to get dx/x , the easy way is "flip" both sides
then move the du to the other side.

Answer 58

in other words, we know how to integrate dx/x and not x/dx

Answer 59

ok

Answer 60

\[\frac{ (2u+1) }{-2-2u^2 }du=\frac{ dx }{ ? }\]

Answer 61

Affirmative?

Answer 62

why the question mark where the x should be ?

Answer 63

there should be an x

Answer 64

\[ \frac{ (2u+1) }{-2-2u^2 }du=\frac{ dx }{ x } \]
we can integrate both sides, but obviously, the right side is easier
just to "clean up" the left side I would factor -2 from the denominator:
\[ - \frac{1}{2}\frac{ (2u+1) }{u^2+1 }du=\frac{ dx }{ x } \]

Answer 65

we have, so far
\[ - \frac{1}{2} \int \frac{2u\ du}{u^2+1} - \frac{1}{2} \int \frac{du}{u^2+1} = \int \frac{dx}{x}\]

Answer 66

arctg?

Answer 67

twice?

Answer 68

the first is dw/w with w= u^2+1

Answer 69

so ln(u^2+1)*-0.5

Answer 70

for the first term

Answer 71

and then -0.5arctgu

Answer 72

yes

Answer 73

but the answer is \[\ln(x^2+y^2)+\arctan(y/x)=c\]

Answer 74

once you get the integrated expression, you can twist it into many different forms with your algebraic skills

Answer 75

wow thanks allot! @phi you always help me with so much patience

Answer 76

thank you!

Answer 77

It does not look particularly easy to get your book answer. Give it a try.

Answer 78

I shall

Answer 79

@phi?

Answer 80

I still don't see how do i get the answer....

Answer 81

@rvc is this as the answer?

Answer 82

ln(x2+y2)+arctan(y/x)=c that's the answer

Answer 83

but we got smth quite different

Answer 84

First, write down what you got after integrating, including the constant of integration

Answer 85

I got \[\frac{ -1 }{ 2 }\ln(\frac{ y }{ x })^2-0.5\arctan(\frac{ y }{ x })=lnx+c\]

Answer 86

is that correct?

Answer 87

now look at what you are trying to get, namely
ln(x2+y2)+arctan(y/x)=c

it's clear we want all the terms except the constant on one side
so move the ln x to the left side

Answer 88

ok

Answer 89

but still

Answer 90

it's not the same

Answer 91

btw, how did you get that ln term
it started as
-1/2 ln (u^2 +1)
with u= y/x

Answer 92

U=y/X

Answer 93

no?

Answer 94

yes, u =y/x . what is u^2 + 1?

Answer 95

it's the integral

Answer 96

\[\int\limits \frac{ dx }{x}=lnx\]

Answer 97

I mean, what is the expression u^2 + 1 after subbing in y/x for u