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

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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 h(t) \geq 0 for all t. Unfortunately the general solution of that particular ODE is oscillatory with a decaying amplitude, so there will be intervals where h(t) < 0 unless h(t) = 0 for all t.
 

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