Can Bernoulli's Equation Solve This Differential Problem?

  • Thread starter abuder3
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In summary, the task is to solve a given equation. No further information or context is provided.
  • #1
abuder3
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any one solve this eq. please ?

hi all
any one solve for me this eq.

y(2xy+ex)dx-exdy=0 y(0)=2
 
Last edited:
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  • #2
Welcome to PF!

Hi abuder3! Welcome to PF! :smile:
abuder3 said:
hi all
any one solve for me this eq.

y(2xy+ex)dx-exdy=0

ah … that's not the way this forum works. :wink:

First show us what work you've done …

which category do you think this comes in?

have you seen anything similar?

what have you tried? :smile:
 
  • #3


:smile::smile::smile::smile:
 
  • #4


abuder3 said:
:smile::smile::smile::smile:

Classic.
 
  • #5


abuder3 said:
hi all
any one solve for me this eq.

y(2xy+ex)dx-exdy=0 y(0)=2
Write it out in the form dy/dx + ... = ..., rearrange the terms and you should be able to get a recognisable DE.
 
  • #6


I can't get any solutions
 
  • #7


please write the solution
 
  • #8


If you click on the "Go advanced" button at the bottom of your reply box it will present you with a number of options to add extra content into your replies.

There is a "[tex]\sum[/tex]" button which you can click on to add mathematics in LaTeX format. Its very useful when trying to communicate ideas to other as ASCII is insufficient for communicating mathematics.

Good luck
:cool:
 
  • #9
abuder3 said:
please write the solution

Show us what work you've done

which category do you think this comes in?

have you seen anything similar?

what have you tried? :smile:
 
  • #10


thanks!
 
  • #11


Hi there!

It's an interesting DE! To be honest I don't have any idea how to solve it too :(

Write it out in the form dy/dx + ... = ..., rearrange the terms and you should be able to get a recognisable DE.

[tex]y(2xy+e^x)dx=e^xdy[/tex]

[tex]\frac{dy}{dx}=\frac{2xy^2+ye^x}{e^x}[/tex]

[tex]\frac{dy}{dx}-y=\frac{2xy^2}{e^x}[/tex]

Now, as I stare at it I see it's not linear and it's not homogeneous. It also cannot be solved by separation of variables.., quadrature does not work here. The 2nd power of y reminds me of the Bernoulli equation.. hmmmmm
 
  • #12


Marin said:
Now, as I stare at it I see it's not linear and it's not homogeneous. It also cannot be solved by separation of variables.., quadrature does not work here. The 2nd power of y reminds me of the Bernoulli equation.. hmmmmm

The latter is correct, consider Bernoulli...
 

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