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

[cocopops12]

(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?

 

[rbj]

if you make that substitution, it's linear.

 

[Ratch]

cocopuff,

Is this DE linear?

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(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

 

[pasmith]

(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.