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another O.D.E. question:
xyy`= 2y^2 +4x^2
also unseparable...
xyy`= 2y^2 +4x^2
also unseparable...
Galileo said:whenever you have y'=F(y/x), the equation also called homogeneous (different from the homogeneous linear ODE definition), this one is invariant under 'zoom'.
Whoopsie, you've lost an x.asdf1 said:hmmm... here's my work~
(i forgot to include that the original condition is y(2)=4, sorry about that!)
y`= 2y/x+4x/y
suppose y=vx
=>v=y/x
so y`=v+v`x
=> v+xv`=2v +4/v
=> v`x=v+4/v
=> v`x=(v^2 +4)/v
=> v/(v^2 +4)dv=dx
The actual answer also needs a square root since you'll get a relation between y^2 and x. Now if this equation is true both the positive and negative square roots must be valid solutions to the ODE, but a little bit of thought shows that no more than one can fit the same boundary condition (y(2)=4)now my problem is that when you take the square root on both sides of the equations, the right side should have a positive and negative root, but the correct answer should only have a positive square root~
why?
A linear DE of, say, second order:btw, what is meant by the different homogeneous definitions and "zoom"?