Is this correct??

[DanielBW]

Hello everyone !! I have this problem: "DEMOSTRATE that the differential equation $xdy-ydx=tan^{-1}(y/x)dx$ can be solved by using the substitution of $y=vx$ So i proceed to solve it:

y=vx

Then

dy = vdx + xdv

And so, for the original differential equation:

$\displaystyle x(vdx+xdv) - xvdx = tan^{-1}(\frac{vx}{x})dx$

Simplifiyng and re-ordening i get:

$\displaystyle x^2dv-tan^{-1}(v)dx=0$ (Separable differential equation)

Solving:

$\displaystyle \int \dfrac{1}{tan^{-1}(v)}dv=\int \dfrac{1}{x^2}dx$

$\displaystyle \int \dfrac{1}{tan^{-1}(v)}dv=-\dfrac{1}{x}+C$

But to complete the problem i need return to the original variable $y$ but i dont know how to solve the left-side integral.... perhaps i'm doing something wrong... can someone help me please???

 

[Ishuda]

Assuming I am reading your problem correctly, you need the solution to


There are no results in standard mathematical functions in WolframAlpha.

 

[DanielBW]

Assuming I am reading your problem correctly, you need the solution to
View attachment 4506

There are no results in standard mathematical functions in WolframAlpha.

Yeah!! that's my problem... i can't solve that integral... so i supose leave it like that and argue that the differential equation only has a implicit solution? By the way sorry for the code... I'm new in this forum and didn't know how fix it...

 

[Ishuda]

Yeah!! that's my problem... i can't solve that integral... so i supose leave it like that and argue that the differential equation only has a implicit solution? By the way sorry for the code... I'm new in this forum and didn't know how fix it...

Your code was fine, I just sometimes misread. If you like, you could try some LaTex,
ftp://ftp.ams.org/pub/tex/doc/amsmath/short-math-guide.pdf
but that is just to make things 'fancy'

Yes, I would leave it as an implicit function as an answer to an exercise. If actual values were wanted, one could do expansions of the arc-tangent function such as at
https://ccrma.stanford.edu/~jos/pasp/Arctangent_Series_Expansion.html
for x 'around zero' to get
ln(x) + $\displaystyle \frac{1}{3}x^2-\frac{1}{45}x^4 + ...$

 

[HallsofIvy]

The problem does NOT say "solve the differential equation...", it says "demonstrate that the differential equation can be solved ... " Showing that it can be reduced to an integral is sufficient.