another O.D.E. question:
xyy`= 2y^2 +4x^2
I`m a bit rusty on D.E.'s, but you can write it as:
IIRC, 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'.
If you make the subsitution z=y/x, the equation becomes:
which is separable.
it's separable, use v=y/x
why'd you think of v=y/x?
Galileo explained. It's because the DE is homogeneous (not with the 0 constant, but with the sum of all exponents of all terms is constant (here 2). You can then divide so the DE becomes a function of (y/x). Susbtitution will make it separable.
so the first step when you see an O.D.E you have to check if it's separable
=> if not, then check if it's homogeneous, and use that substitution...
Checking if it's separable should indeed always be the first you do and when it's not, try to see if you can make it separable with a substitution.
The substitution used here a very common one.
what are the most common substitutions used?
Well y = v(x)x is a common one and if I recall correctly, for second order isobaric equations you can make the substitution y = vx^m if there is a single value of m which makes the equation dimensionally consistent (you assign weights to y, dy, x and dx).
ok~ thanks!!! :)
hmmm... here's my work~
(i forgot to include that the original condition is y(2)=4, sorry about that!)
=> v+xv`=2v +4/v
=> v`x=(v^2 +4)/v
=> v/(v^2 +4)dv=dx
=> 1/2[dv^2/(v^2 +4)] =dx
=> 1/2ln[absolute value (v^2 +4)] =x+ c`
=> ln[absolute value (y^2/(x^2) +4)]=2x+c`
=> y^2/(x^2) +4 =e^(2x +c`)
=> y^2 +4x^2 =c(x^2)*[e^(2x)]
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~
btw, what is meant by the different homogeneous definitions and "zoom"?
Whoopsie, you've lost an x.
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)
A linear DE of, say, second order:
is called homogeneous if r(x)=0. Your ODE is a also called homogeneous, but it's in a different context, so it naturally hasn't the same meaning.
Since the ODE can be written in terms of y/x. If we make the change of variables [itex]\hat y=ay[/itex] and [itex]\hat x=ax[/itex] for some positive a. So that we're scaling the y and x-axes by the same amount (this is what I meant by zoom), the ODE looks the same, since [itex]\hat y/\hat x=y/x[/itex].
thank you very much!!! :)
i'll try and be more careful on my calculations~
Separate names with a comma.