Solve DE Linearity: y^2 + (y^2)' + y^2 = 0

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Discussion Overview

The discussion revolves around the linearity of the differential equation (DE) given by (y^2)'' + (y^2)' + y^2 = 0. Participants explore the implications of substituting y^2 with a new variable h and the validity of this approach in terms of linearization.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants question whether the DE is linear, with one asserting that it is not due to the dependent variable being "y" and y^2 being nonlinear.
  • Others propose that substituting y^2 = h and solving for h would result in a linear equation.
  • There is a discussion on the validity of the substitution, with some affirming it as valid.
  • Concerns are raised about the classification of linear DEs, with one participant stating that they are not categorized as trivial or nontrivial.
  • One participant notes that for y to be real-valued, h(t) must be non-negative, and mentions the oscillatory nature of the general solution, indicating potential issues with h(t) being negative.

Areas of Agreement / Disagreement

Participants express differing views on the linearity of the DE and the implications of the substitution, indicating that multiple competing views remain without a clear consensus.

Contextual Notes

There are limitations regarding the assumptions made about the substitution and the conditions under which h(t) remains non-negative. The discussion does not resolve the implications of these assumptions.

cocopops12
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(y^2)'' + (y^2)' + y^2 = 0

1) Is this DE linear?

What if we substitute y^2 = h
and solve for h
y = sqrt(h)

2) Would that be valid?
3) Would that be considered a somewhat trivial type of linearization?
 
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if you make that substitution, it's linear.
 
cocopuff,

Is this DE linear?

--------------------------------------------------------------------------------
(y^2)'' + (y^2)' + y^2 = 0

No, of course not. The dependent variable is "y" and y^2 is nonlinear.

What if we substitute y^2 = h
and solve for h
y = sqrt(h)

2) Would that be valid?

Yes.

3) Would that be considered a somewhat trivial type of linearization?

No, linear DE's are not classified as trivial linear or nontrivial linear DE's.

Ratch
 
cocopops12 said:
(y^2)'' + (y^2)' + y^2 = 0

1) Is this DE linear?

What if we substitute y^2 = h
and solve for h
y = sqrt(h)

2) Would that be valid?

Yes.

If you want y to be real-valued, then you will need [itex]h(t) \geq 0[/itex] for all t. Unfortunately the general solution of that particular ODE is oscillatory with a decaying amplitude, so there will be intervals where [itex]h(t) < 0[/itex] unless [itex]h(t) = 0[/itex] for all t.
 

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