I really dont know what to do whit this equation, i have tried to integrate it, but i can do it.y′3−yy′2−x2y′+x2y=0

Name: I really dont know what to do whit this equation, i have tried to integrate it, but i can do it.y′3−yy′2−x2y′+x2y=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: I really dont know what to do whit this equation, i have tried to integrate it, but i can do it.y′3−yy′2−x2y′+x2y=0

Question

Cheryl Stark · Accepted Answer

One thing which can be done with the equationy′3−yy′2−x2y′+x2y=0    (1)is to observe thaty′3−yy′2−x2y′+x2y=y′2(y′−y)−x2(y′−y)=(y′2−x2)(y′−y)=(y′+x)(y′−x)(y′−y)    (2),so that (1) becomes(y′+x)(y′−x)(y′−y)=0.    (3)It now appears that (1) is really three ODEs in one, as it were, since a solution toy′+x=0,    (4)ory′−x=0    (5)ory;−y=0    (6)will satisfy (1). Taking y(0)=y0 as the initial condition (for example; there are other possibilities y0=y(x0) for x0ϵR), we see that the solutions to (4)-(6) arey=−12x2+y0    (7)y=12x2+y0    (8)andy=y0ex,    (9)respectively, and exactly one of these holds. It seems to me that establishing this point rigorously will involve invoking smoothness and continuation arguments, but I&#039;m leaving such an in-depth study to my readers. From an algebraic point of view, we are choosing a branch of solutions to the cubic equation in y′ (1); there seems to be an interesting if elementary interplay between ideas from algebraic geometry and analysis taking place here, but I haven&#039;t the time right now to be more thorough; I merely present intuitions I have attained by groping in the dark in the hope that someone can take things further.But anyway, thare&#039;s (at least) one thing which can done with equation (1).Hope this helps :)

Hashim Townsend · Accepted Answer

y′3−yy′2−x2y′+x2y=0Factoring,y′2(y′−y)−x2(y′−y)=0(y′−y)(y′2−x2)=0(y′−y)(y′+x)(y′−x)=0This will give you three solutions.

I really dont know what to do whit this equation,

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