2012-09-13, 06:45 PM
Rewrite:
(y'')^2 -y(y'') +(y^2)y' = 0
Use quadratic formula:
y'' = [y ±√(y^2 -4(y^2)y'] /2
Simplify:
(2y''-y)^2 = y^2 (1 -4y')
Note: At this stage, it is clear that if y=0, then 2y''-y =0, so y''=0. This corresponds to the trivial
solution of y=0.
Simplify further:
(2y''/y^2 -1)^2 = 1 -4y'
At this stage, note that the equation holds when both sides equal the same constant.
(2y''/y^2 -1)^2 = 1 -4y' = C
You end up with two easier equations:
A) 1 -4y' = C
B) 2y''- (1±√C)*y^2 = 0
(y'')^2 -y(y'') +(y^2)y' = 0
Use quadratic formula:
y'' = [y ±√(y^2 -4(y^2)y'] /2
Simplify:
(2y''-y)^2 = y^2 (1 -4y')
Note: At this stage, it is clear that if y=0, then 2y''-y =0, so y''=0. This corresponds to the trivial
solution of y=0.
Simplify further:
(2y''/y^2 -1)^2 = 1 -4y'
At this stage, note that the equation holds when both sides equal the same constant.
(2y''/y^2 -1)^2 = 1 -4y' = C
You end up with two easier equations:
A) 1 -4y' = C
B) 2y''- (1±√C)*y^2 = 0