First Order differential equation

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ViliusP2

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Hello everyone, I am having a problem with this differential equation:
, where a, b, c and k are constants. I don't want you to give me the full answer, but I wish to know the method by which such equation should be solved. Right now I am completelly lost, as I don't have a slightest Idea, how I should start solving this equation.
 
ViliusP2 said:
Hello everyone, I am having a problem with this differential equation:
, where a, b, c and k are constants. I don't want you to give me the full answer, but I wish to know the method by which such equation should be solved. Right now I am completelly lost, as I don't have a slightest Idea, how I should start solving this equation.
Click to expand...

Is your problem:

\(\displaystyle \displaystyle{\left (c-k*t-a^{-b*y(x)}\right ) * \left (\dfrac{dy}{dx}\right )^2 - \dfrac{dy}{dx}} \ = \ 0\)

are you sure of the terms 'kt' and (dy/dx)?
 
Last edited by a moderator:
Thanks for replying, but my problem is little different:
[FONT=MathJax_Math]c[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Size2])[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Size3]([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Size3])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x)[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0 and yes, I am sure about those terms. Sorry about the equation format, I do not know yet, how to write them in this forum.
Also, It would be nice if you gave me advice, on what numerical method would be best to solve this equation, if there is no analyticall method.[/FONT]
 
ViliusP2 said:
Thanks for replying, but my problem is little different:
[FONT=MathJax_Math]c[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]t[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Size2])[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Size3]([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Size3])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x)[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0 and yes, I am sure about those terms. Sorry about the equation format, I do not know yet, how to write them in this forum.
Also, It would be nice if you gave me advice, on what numerical method would be best to solve this equation, if there is no analyticall method.[/FONT]
Click to expand...

Assuming 't' is time and not a constant - y becomes function of both x and t. In that case - shouldn't those "differentials" be "partial differential"?
 
Oh sorry, my mistake. It is not kt actually but kx, therefore whole equation is:
[FONT=MathJax_Math]c[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Size2])[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Size3]([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Size3])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x)[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0[/FONT]
 
ViliusP2 said:
I do not know yet, how to write [math expressions] in this forum.
Click to expand...

ViliusP2 said:
[FONT=MathJax_Math]c[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]k[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Math]b[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Size2])[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Size3]([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Size3])[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]−([/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]y/[/FONT][FONT=MathJax_Math]d[/FONT][FONT=MathJax_Math]x)[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]0
[/FONT]
Click to expand...

Hi ViliusP2:

The equation above may also be texted as

c - kx - a^(-by)*(dy/dx)^2 - dy/dx = 0


You may find information at this page helpful.

Cheers :cool:
 
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