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

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In summary, when substituting y^2 = h and solving for h, the resulting solution is valid and considered a nontrivial linearization of the given differential equation. However, in order for the solution to be real-valued, the function h(t) must be greater than or equal to 0 for all t. The general solution of the ODE is oscillatory with a decaying amplitude, so there may be intervals where h(t) is negative unless h(t) is equal to 0 for all t.
  • #1
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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  • #2
if you make that substitution, it's linear.
 
  • #3
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
 
  • #4
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.
 
  • #5


1) Yes, this DE is linear because it only contains first and second derivatives of the dependent variable y, and there are no higher powers or products of y present.
2) Substituting y^2 = h is a valid method to solve this DE, as it simplifies the equation and allows us to use methods for solving linear DEs.
3) This can be considered a trivial linearization because it involves a simple substitution and does not require more complex techniques such as variation of parameters or integrating factors. However, it is still a valid and effective method for solving the DE.
 

What is a differential equation (DE)?

A differential equation is a mathematical equation that relates a function with its derivatives. It is commonly used to describe physical phenomena and can be solved to determine the behavior of the function.

What does it mean for a DE to be linear?

A linear differential equation is one in which the dependent variable and its derivatives appear only in a linear combination. In other words, the function and its derivatives are only multiplied by constants or added together.

How do you solve a linear DE?

To solve a linear differential equation, you can use various methods such as separation of variables, integrating factors, or the method of undetermined coefficients. In this particular equation, you can use the method of undetermined coefficients to determine the solution.

What is the solution to y^2 + (y^2)' + y^2 = 0?

The solution to this linear differential equation is y = 0. This can be determined by using the method of undetermined coefficients and setting y = Ce^(-x), where C is a constant. By substituting this into the equation, we get C = 0, thus giving us the solution y = 0.

What is the importance of solving DEs in scientific research?

DEs are used in various fields of science to model and understand real-world phenomena. By solving DEs, scientists can make predictions and analyze the behavior of systems. They are also crucial in the development of technologies such as engineering, physics, and biology.

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