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I wrote up it as a first order diff.equation with $2$ variables by introducing $x:=y"$, so that we are looking for (the $y$-coordinate of) the trajectory of the vector field $(1/y^3, x)$, I could draw some vectors, & could "guess" how the solution will look lượt thích.
Look for the solution in form $y=c,t^alpha$. I got that $alpha=1/2$ và $c^4=-4$, that is $c=pm1pm i$. However, I guess, real function was asked.
I guess I"m missing something I should know about solving such a differential equation.
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asked May 31 "13 at 7:41
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Multiply $y"$,$$y"y""=y^-3y"$$Integrate, $$frac 1 2 (y")^2 = -frac 1 2 y^-2 + c$$Using initial condition, $$2=-frac 1 2 +c$$Thus $c=frac 5 2$. Thus, we have $$(y")^2+y^-2=5.$$Multiply $y^2$, and substitute $Y=y^2$, then$$frac (Y")^2 4 +1 = 5Y,$$$$(Y")^2=20Y-4$$I think it is easy to go from here.
answered May 31 "13 at 7:59
Sungjin Kyên ổn
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