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NOnlinear differential equation method help?
I don't get this method for nonlinear ODES
example F(y,p)=0 y=ø(p) then dx=ø'(p)dp/p
x=Integral [ ø'(p)dp/p
another example is
x=p+sin(p/x) where p=y'=xt <=why??
then
x=sin(t)/1-t
p=t*sin(t)/1-t
dy=t sin(t) *[(1-t)cost+sint]/[1-t^3] dt <= where does all of this come from????????
so then we would integrate for y? what is this method with parameters I don't get it. Can I read about it somewhere?
1 Answer
- ?Lv 78 years agoFavorite Answer
Re your second example: I think the original problem must be to solve
y'(x) + sin[y'(x)/x] = x
Let y' = p merely for the simplicity of writing
x = p + sin(p/x)
The parameter t is introduced with the aim of arranging a single expression for a derivative and we assume, (guess), that dy/dx can be written in the form xt(x) and write
p = xt
x - p = x(1 - t) = sin(p/x) = sin(t)
x = sin(t)/(1 - t)...................(a)
so that now we have
dy/dx = p = [t/(1 - t)]sin(t)....(b)
but we want dy/dt = dy/dx*dx/dt, so we need to calculate dx/dt from (a)
we use quotient rule: [u/v]' = (vu' - uv')/v^2
dx/dt = [(1 - t)cos(t) + sin(t)]/(1 - t)^2...(c)
Multiplying (b) by (c) yields
dy/dt =t sin(t)[(1 - t)cost + sint]/(1 - t)^3
[Note: not [1 - t^3] in denominator}
That explains the part you asked about .
Integration is messy, but just possible (Wolfram) leading to a result
y = f(t) that uses special integral functions Ci(t) and Si(t)
There remains the unexplained part of how to use y = f(t) to get to
y = f(x). Perhaps your teacher will elaborate at your next class.
Regards - Ian