Question

how would you solve this differential equation...

Answer 1

\[ycos(x/y) + [ysin(x/y) - xcos(x/y)] y \prime = 0\]

Answer 2

?

Answer 3

@amistre64

Answer 4

i dont know how to use the substitution.

Answer 5

I tried but lose the dv/dx somewhere along my steps.

Answer 6

Any ideas guys?!

Answer 7

I used v = x/y to simplify.

Answer 8

Then solved for y.

Answer 9

Getting \[y= x/v\]

Answer 10

Then I went to get \[y\]

Answer 11

I mean \[y\]

Answer 12

wow...i mean \[\frac{ dy }{ dx}\]

Answer 13

@amistre64 any idea?

Answer 14

@electrokid

Answer 15

@hartnn

Answer 16

@kropot72

Answer 17

@Preetha

Answer 18

@radar

Answer 19

after simplification, i am getting
y cos v dv/dy = -sin v

Answer 20

lets start with step 1
y cos v = - [y sin v - vycos v]y' ......[note y cancels]

x=vy
so, dx/dy = v+ y dv/dy

Answer 21

cos v (v+y dv/dy) = v cos v - sin v
got this much ?

Answer 22

@qrious126
then it will be variable separable.

Answer 23

-cot v dv = 1/y dy
easily integrable.

Answer 24

Don't mind me, just writing it in a more me-friendly manner...
\[\large y\cos\left(\frac{x}y\right)+\left[y\sin\left(\frac{x}y\right)-x\cos\left(\frac{x}y\right)\right]\frac{dy}{dx}=0\]

Answer 25

Owl be back...
Owl be here...
Owl go crazy if I see that owl again in that fashion >.<

Anyway
\[\large y\cos\left(\frac{x}y\right)\color{blue}{dx}+\left[y\sin\left(\frac{x}y\right)-x\cos\left(\frac{x}y\right)\right]{dy}=0\]

\[\large y\cos\left(\frac{x}y\right){dx}+y\sin\left(\frac{x}y\right)\color{blue}{dy}-x\cos\left(\frac{x}y\right)\color{blue}{dy}=0\]

\[\large y\cos\left(\frac{x}y\right){dx}\color{green}{-x\cos\left(\frac{x}y\right){dy}+y\sin\left(\frac{x}y\right){dy}}=0\]

\[\large \color{red}{\cos\left(\frac{x}y\right)\left(y{dx}-x{dy}\right)}+y\sin\left(\frac{x}y\right){dy}=0\]

\[\large \cos\left(\frac{x}y\right)\left(\frac{y{dx}-x{dy}}{\color{green}{y^2}}\right)+\frac{\sin\left(\frac{x}y\right){dy}}{\color{green}y}=0\]

\[\large \cos\left(\frac{x}y\right)\left(\color{red}{d\frac{x}{y}}\right)+\frac{\sin\left(\frac{x}y\right){dy}}{y}=0\]

\[\large \text{Let }u=\frac{x}y\]

\[\large \cos\left(\color{red} u\right)\left(\color{red}{du}\right)+\frac{\sin\left(\color{red}u\right){dy}}{y}=0\]


And the rest is history, I guess...

Answer 26

Let me Latexify it also.
\( \large \dfrac{x}{y} = v, so x= vy \)
differentiating w.r.t y,
\( \large \dfrac{dx}{dy} = v+y \dfrac{dv}{dy}
\\ \large \large y\cos\left(\frac{x}y\right)+\left[y\sin\left(\frac{x}y\right)-x\cos\left(\frac{x}y\right)\right]\frac{dy}{dx}=0
\\ \large y\cos\left(\frac{x}y\right)\frac{dx}{dy}=-\left[y\sin\left(\frac{x}y\right)-x\cos\left(\frac{x}y\right)\right]
\\ \large cos v(v+y\dfrac{dv}{dy})=-sin v+vcos v
\\ \large \cancel{v cos v}+ycos v \dfrac{dv}{dy}= -sin v +\cancel{v cos v}
\)
Separating the variables
\( \large -\dfrac{cos v}{sin v} dv=\dfrac{1}{y}dy
\\ \large –cot v dv =\dfrac{1}{y}dy
\)
Now integrate both sides, you can solve further, right ?

Answer 27

Latexify... that's one for the journal...

Answer 28

you got 2 methods, follow whichever you can understand :)

Answer 29

I was able to follow your method. Thank you!

Answer 30

By the ways, thank you all!

Answer 31

welcome ^_^
glad we could help :)

Answer 32

I am getting \[y= \sin(x/y)\].

Answer 33

The funny thing is I get the differential equations concepts, but find myself getting stuck with the simple algebra and trig. Silly me!

Answer 34

:)
^all I could muster
OS went all "owl be back" while I was typing that lengthy thing back there :/

Answer 35

haha

Answer 36

-cot v dv = 1/y dy
-log sin v dv = log y +log c
log (1/sin v) = log (yc)
1/sin (x/y) = yc
how you got sin instead of 1/sin ??

Answer 37

or y sin (x/y) =c

Answer 38

I caught my arithmetic error. Thanks for pointing that out!

Answer 39

welcome :)

Answer 40

shouldnt you answer be:\[y/\sin (x/y) = c?\]

Answer 41

your*

Answer 42

you got this ?
then multiplying sin (x/y) on both sides.

Answer 43

**1/sin (x/y) = yc

Answer 44

1/c is still a constant, got you!

Answer 45

oh, yes that., yes.

Answer 46

Thanks!

Answer 47

how should I attempt to simplify this equation:\[((1/y) - x^3\sin^2(x)\ln(y))/(1+(x^3/y)\sin^2(x))\] ?

Answer 48

It relates to exact equations.

Answer 49

The original problem is:\[(1+(x^3/y)\sin^2(x))dx + ((1/y)(x+(1/\cos^2(2y))dy = 0\]

Answer 50

I am using \[(Nx - My)/M\]

Answer 51

since the partial fractions are not exact.