Is there a general way of solving ODE's of the form f(y,y')=0?

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The discussion focuses on solving ordinary differential equations (ODEs) of the form f(y, y') = 0, specifically through the method of treating "a" as a parameter and deriving functions t = g(a). The process involves substituting t with y' and a with y, leading to the equation dy/dx = g(y). An example provided is y² + y'² - 1 = 0, which simplifies to x = arcsin(y) + C, yielding y = sin(x - C). Challenges include the inability to analytically solve f(a, t) = 0, lack of a known primitive for 1/g(x), and difficulties in computing the reciprocal of x = f(y).

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Is there a general proposed way of solving ODE's of the form f(y,y')=0? any ideas?
 
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AdrianZ said:
Is there a general proposed way of solving ODE's of the form f(y,y')=0? any ideas?

Solve the equation f(a,t)=0 for t, considering "a" as a parameter. The result is one ore several functions t=g(a)
Let t=y' and a=y
y'=g(y)
For each function g :
dy/dx = g(y)
integrate (1/g(y))dy = dx
the result is on the forme x = f(y)
calculate x = reciprocal function of f(y)

Example :
y²+y'²-1 = 0
equation to be solved : f(a,t) = a²+t²-1 = 0
t = sqrt(1-a²)
dy/dx = sqrt(1-y²)
dx = dy/sqrt(1-y²)
x = arcsin(y) +C
y = sin(x-C)

However, some difficulties might be encountered :
- If analytical solving of equation f(a,t)=0 is not possible.
- if a primitive of the function 1/g(x) is not known
- if the reciprocal of function x=f(y) cannot be analytically computed.
 

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