We have
dx/dt =2v*(-x) / sqrt((-x)^2+(vt-y)^2)
&
dy/dt = 2v*(vt-y) / sqrt((-x)^2+(vt-y)^2)
we can get dy/(vt-y) = -dx/x with some math, and when I integrate both sides, I get
vt-y = x*e^c
and the question asks for y
so, the answer would be y = vt-x*e^c wouldn't it?
My question comes...
The direction of N at time t would be the vector from N to M, which is
the vector from (x,y) to (0,vt) = -x.i + (vt-y).j
I think, dx would be 2v*(x component of the direction vector) which is
2v*(-x)
respectively, dy would be 2v*(vt-y) ?
Hello again,
I come up with a new question which I can't express as a differential equation form.
2 cars start on x axis, M at the origin and N at the point (36,0)
Suppose:
M moves along the y axis,
and N moves directly toward M at all times,
and N moves twice as fast as M.
Q) How far will...
Re: First Order Non Linear Ordinary D.E.
Indeed, even wolfram can't handle this problem.
I'm trying to use WolframMathematica but I can not make it understand the question because I'm not familiar with it...
There are no hints, suggestions related to the question :(
Re: First Order Non Linear Ordinary D.E.
Yes Sudharaka solved a different D.E
This is the exact problem with the corrected parantheses..
(x*y*sqrt(x^2-y^2) + x)*y' = (y - x^2*(sqrt(x^2-y^2)))
Thank you in advance
Hello people,
I couldn't solve the given D.E by using exact d.e & substitution method :(
Thanks in advance.
(x*y*sqrt(x^2-y^2) + x)*y' = (y - x^2*(sqrt(x^2-y^2) )
gif file of d.e can be found in the attachments part.