1. ## variable substitution question(diff)

$\displaystyle \frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\$
$\displaystyle t=\arctan y\\$
$\displaystyle t'=\frac{1}{1+y^2}y'\\$
$\displaystyle \frac{x}{\ln (x)t-1}=\frac{t}{t'}\\$

$\displaystyle \frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\$
$\displaystyle xdt=(\ln (x)t-1)tdx\\$
$\displaystyle \frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\$
still cant beak it as one type of variable on each side
so i substitute by another variable
$\displaystyle z=\ln x\\$
$\displaystyle dz=\frac{dx}{x}\\$
$\displaystyle \frac{dt}{zt-1}=dz\\$
i dont know how to separate each variable type on one side
so i could integrate

??

2. Hi transgalactic

Do you know integrating factor ?

3. yes i know that.but how did you know that that i should solve it like this in here?

4. Originally Posted by transgalactic
[snip]

$\displaystyle \frac{xy'}{(\ln x\arctan y)-1}=(1+y^2)\arctan y\\$
$\displaystyle t=\arctan y\\$
$\displaystyle t'=\frac{1}{1+y^2}y'\\$
$\displaystyle \frac{x}{\ln (x)t-1}=\frac{t}{t'}\\$

$\displaystyle \frac{x}{\ln (x)t-1}=\frac{tdx}{dt}\\$
$\displaystyle xdt=(\ln (x)t-1)tdx\\$
$\displaystyle \frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\$

[/snip]
From here,

$\displaystyle \frac{\,dt}{t\ln x-1}=\frac{\,dx}{x}\implies \frac{\,dt}{\,dx}=\frac{\ln x}{x}t-\frac{1}{x}$ $\displaystyle \implies \frac{\,dt}{\,dx}-\frac{\ln x}{x}t=-\frac{1}{x}\sim \frac{\,dt}{\,dx}+P(x)t=Q(x)$.

Can you take it from here and apply the integrating factor technique (as songoku mentioned)?

5. ok i rearanged it like i used to in order to find integration factor
$\displaystyle xdt=(tlnx-1)dx$
i got that the integration factor is:
$\displaystyle y=e^{\frac{ln^2x}{2}-lnx}$
so i mltiply both sides by y
$\displaystyle xe^{\frac{ln^2x}{2}-lnx}dt=(tlnx-1)e^{\frac{ln^2x}{2}-lnx}dx$
now its even more complicated than before
??

6. Hi transgalactic

Sorry, I have tried to solve this and ended up finding the integration that can't be stated in elementary function.
For now, maybe we can wait for Chris L T521's help

7. this is a fll solution:
http://i30.tinypic.com/2lllrpk.gif

its really frustrating that i did all the things by the book and i didnt manage to solve its
what is the mistake in diagnozing the wa of solution

what should step there in my equation is a red light to solve it like
the solution above?

8. Hi transgalactic

I re-check your work and find out the mistake.

$\displaystyle xdt=(\ln (x)t-1)tdx\\$

$\displaystyle \frac{dt}{\ln (x)t-1}=\frac{dx}{x}\\$ -----------> you missed the term "t"

So, that's not linear form anymore but Bernoulli form

9. $\displaystyle \frac{dt}{(\ln (x)t-1)t}=\frac{dx}{x}\\$
$\displaystyle z=\ln x$
$\displaystyle \frac{dt}{(zt-1)t}=dz\\$

i dont have here a bernuly equation ..
what to do?
bernule needs to have f'() form
i dont know how transform it to this form

10. Hi transgalactic

$\displaystyle \frac{dt}{(zt-1)t}=dz\\$

$\displaystyle \frac{dt}{dz}=zt^2-t$

$\displaystyle \frac{dt}{dz}+t=zt^2$

11. here is another problem:
all those variable are confusing
how did you know to t it in this way
dt/dz

or

dz/dt

why you put the derivative part in that way

what is the logic behind it
?

12. Hi transgalactic

It's not matter whether you solve $\displaystyle \frac{dt}{dz}$ or $\displaystyle \frac{dz}{dt}$ because it will give you the same result.

If you state it in $\displaystyle \frac{dz}{dt}$ , it will be :

$\displaystyle \frac{dz}{dt}=\frac{1}{zt^2-t}$

Can you solve it ?

13. the problem is that i cant see what variable is a function of what variable
if i had dy and dx
then i whould do
dy/dx

but here dz and dt
i dont know who is more superior

14. Originally Posted by Chris L T521
From here,

$\displaystyle \frac{\,dt}{t\ln x-1}=\frac{\,dx}{x}\implies \frac{\,dt}{\,dx}=\frac{\ln x}{x}t-\frac{1}{x}$ $\displaystyle \implies \frac{\,dt}{\,dx}-\frac{\ln x}{x}t=-\frac{1}{x}\sim \frac{\,dt}{\,dx}+P(x)t=Q(x)$.

Can you take it from here and apply the integrating factor technique (as songoku mentioned)?
The integrating factor is $\displaystyle \varrho\left(x\right)=e^{-\int\frac{\ln x}{x}\,dx}=e^{-\frac{\ln^2\left(x\right)}{2}}$.

So it follows now that $\displaystyle \frac{\,d}{\,dx}\left[te^{-\frac{\ln^2\left(x\right)}{2}}\right]=-\frac{1}{x}e^{\frac{-\ln^2\left(x\right)}{2}}$

Thus, $\displaystyle te^{-\frac{\ln^2\left(x\right)}{2}}=-\int\frac{1}{x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx\implies t=-e^{\frac{\ln^2\left(x\right)}{2}}\cdot\int\frac{1} {x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx$.

Therefore, $\displaystyle \arctan y=-e^{\frac{\ln^2\left(x\right)}{2}}\cdot\int\frac{1} {x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx\implies y=\tan\left[-e^{\frac{\ln^2\left(x\right)}{2}}\cdot\int\frac{1} {x}e^{\frac{-\ln^2\left(x\right)}{2}}\,dx\right]$

It looks weird, but I guess that's the solution you're looking for...

15. Hi Chris L T521 and transgalactic

@ Chris L T521:
There is error in OP's work.

@transgalactic :
Example : $\displaystyle \frac{dy}{dx}=x$

$\displaystyle dy=x\; dx$

$\displaystyle y=\frac{1}{2}\; x^2 + c$

If you re-arrange the question : $\displaystyle \frac{dx}{dy}=\frac{1}{x}$

$\displaystyle x\; dx = dy$

$\displaystyle \frac{1}{2}x^2+c=y$

You will have same result. So, it doesn't matter if you solve $\displaystyle \frac{dt}{dz}$ or $\displaystyle \frac{dz}{dt}$