Question

Anyone good at Differential equations? That would not mind me monopolizing their time?

Answer 1

I need to find the First integral?<---explain that please. For the exact equation(maybe?)
\[\frac{dx}{z+y}=\frac{dy}{x+z}=\frac{dz}{x+y}\]

Answer 2

@eliassaab Any thoughts?

Answer 3

or @hero ?

Answer 4

Any ideas, even if you don't know are more than welcome

Answer 5

One obvious solution is x=y=z

Answer 6

Have you considered integrating everything? Assuming \(x,y,z\) are independent of each other.

Answer 7

right, but how can we arrive there. And the issue is that would be partials no?

Answer 8

Another one x=C1; y=C2 ;z = C2

Answer 9

ok what if we expressed it as a vector, could we solve it then?

Answer 10

Consider using the chain rule like
\[
\frac{\text{dy}}{\text{dx}}=\frac{\text
{dy} \text{dz}}{\text{dx} \text{dz}}
\]
and try to find a relation between the varialbles

Answer 11

Sorry, swith the the dx and dz in the denominator

Answer 12

it's ok

Answer 13

\[
\frac{\text{dy}}{\text{dx}}=\frac{\text
{dy} \,\text{dz}}{\text{dz}\, \text{dx}}
\]

Answer 14

hmmm.... do you think I could set it up as
\[(x+z)dz=(x+y)dy=(z+y)dz\]

Answer 15

Yes, something like that

Answer 16

oops \[(x+z)dx=(x+y)dy=(z+y)dz\]

Answer 17

That legal? I think it should be

Answer 18

I think so.

Answer 19

Sorry, I have to go now.

Answer 20

it's ok, thank you :)

Answer 21

no one at my university can help with this review question...

Answer 22

@hartnn can you help?

Answer 23

crazy question, eh? @freckles

Answer 24

lol puzzling so far
but I will give it some more attention tomorrow
I know you are studying for exam so that sucks but its also 12:33 am here :(

Answer 25

1:34 here :( But it's alright. I'll just fail haha. The whole class will....

Answer 26

yay for curves!

Answer 27

It is always good to never be alone in failing

Answer 28

true

Answer 29

or at least it makes you feel a little better about failing

Answer 30

also true

Answer 31

http://www.math.utk.edu/~ccollins/M512/HW/soln5.pdf
this first question looks like a similar type of equation

Answer 32

Though our equations look like we get know cancellations

Answer 33

like the one there

Answer 34

no*

Answer 35

\[dx(x+z)=dy(z+y) \\ dy(z+y)=dz(x+z) \\ x+z=\frac{dy}{dx}(z+y) \\ dy(z+y)=dz \cdot \frac{dy}{dx}(z+y) \\ (z+y)=dz \frac{z+y}{dx} \\ 1=\frac{dz}{dx}\]
make sure I didn't make a mistake

Answer 36

1 dx=1 dz
x=z+c

Answer 37

Like so... we can find y now... I suppose

Answer 38

we can replace our x's with (z+c)'s

Answer 39

I love you

Answer 40

you have two x+z and no y+z

Answer 41

\[dx(x+z)=dy(z+y) \\ dy(z+y)=dz(\color\red{x+y}) \\ x+z=\frac{dy}{dx}(z+y) \\ dy(z+y)=dz \cdot \frac{dy}{dx}(z+y) \\ (z+y)=dz \frac{z+y}{dx} \\ 1=\frac{dz}{dx}\]

Answer 42

omg the second line should by (x+y)dy=(z+x)dz right?

Answer 43

(x+z)dx=(x+y)dy=(z+y)dz

Answer 44

I think it should be like that

Answer 45

\[(x+z)dx=(x+y)dy\\(x+y)dy=(z+y)dz\]

Answer 46

goes I keep changing my x's to z' or the other way around

Answer 47

I like that better

Answer 48

gosh*

Answer 49

haha

Answer 50

let me try to fix this

Answer 51

\[\frac{dx}{z+y}=\frac{dy}{x+z}=\frac{dz}{x+y}\]
first to second
\[\frac{dx}{z+y}=\frac{dy}{x+z}\]
\[(x+z)dx=(z+y)dy\]
second to third
\[\frac{dy}{z+x}=\frac{dz}{x+y}\]
\[(x+y) dy=(z+x)dz\]
first to third
\[\frac{dx}{z+y}=\frac{dz}{x+y} \\ (x+y) dx=(z+y) dz\]

Answer 52

i agree

Answer 53

\[(x+y)dy=(z+x) dz \\ (x+y)dx=(z+y) dz \\ (x+y)=(z+x) \frac{dz}{dy} \text{ first equation \in this mini-post } \\ \text{ plug into second equation \in this mini-post } \\ (z+x) \frac{dz}{dy} dx=(z+y) dz \\ (z+x) \frac{dx}{dy}=z+y\\ \text{ but \in the first equation \in my mini-post \above } \\ \text{ the } (z+x) dx=(z+y) dy \\ \ \\\] this didn't help any because we already had that equation

err maybe we can solve those equations for dz above
\[\frac{x+y}{z+x} dy=dz \\ \frac{x+y}{z+y} dx=dz \\ \frac{x+y}{z+x} dy=\frac{x+y}{z+y} dx \\ \frac{1}{z+x} dy=\frac{1}{z+y} dx \]
still don't know what to do with the z

Answer 54

(x+y)dy=(z+x)dz(x+y)dx=(z+y)d< not accurate you mixed up your zs and xs again I think or did something I missed

Answer 55

Yeah I will come back tomorrow
sorry

Answer 56

it's ok, thanks for trying

Answer 57

stay warm and sleep well

Answer 58

One can show z=x + Constant. let us suppose the z=x, then one get that
\[
\frac{\sqrt[3]{\sqrt{e^{6 c_1}-4 e^{3 c_1} x^3}+e^{3 c_1}-2
x^3}}{\sqrt[3]{2}}+\frac{\sqrt[3]{2} x^2}{\sqrt[3]{\sqrt{e^{6 c_1}-4 e^{3 c_1}
x^3}+e^{3 c_1}-2 x^3}}
\]
I got this with Mathematica

Answer 59

\[
y=\frac{\sqrt[3]{\sqrt{e^{6 c_1}-4 e^{3 c_1} x^3}+e^{3 c_1}-2
x^3}}{\sqrt[3]{2}}+\frac{\sqrt[3]{2} x^2}{\sqrt[3]{\sqrt{e^{6 c_1}-4 e^{3 c_1}
x^3}+e^{3 c_1}-2 x^3}}
\]