Solution Verification for Dy = 2y/x using y = Cx^2

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Homework Help Overview

The discussion revolves around verifying whether the function \( y = Cx^2 \) is a solution to the differential equation \( Dy = \frac{2y}{x} \). Participants are exploring the relationship between the left-hand side and right-hand side of the equation.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to differentiate \( y = Cx^2 \) and equate it to the right-hand side of the differential equation. Questions arise about the next steps in the verification process, including whether to substitute \( y \) back into the equation.

Discussion Status

Some participants confirm that the manipulation of the equation leads to a valid equality, while others express uncertainty about the implications of their findings. There is an acknowledgment of the correctness of the derived equality without reaching a definitive conclusion.

Contextual Notes

Participants are working within the constraints of verifying a solution to a differential equation, with some expressing confusion about the verification process and the steps involved.

Dr Game
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Homework Statement



Verify that the function is a solution of the DE

[tex]Dy = \frac{2y}{x} , y = Cx^2[/tex]

2. The attempt at a solution

[tex]LHS = Dy = D(Cx^2) = 2 Cx[/tex]
[tex]RHS = \frac{2y}{x}[/tex]

then... I really don't know what to do from there... do I just simply things?
 
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Dr Game said:

Homework Statement



Verify that the function is a solution of the DE

[tex]Dy = \frac{2y}{x} , y = Cx^2[/tex]

2. The attempt at a solution

[tex]LHS = Dy = D(Cx^2) = 2 Cx[/tex]
[tex]RHS = \frac{2y}{x}[/tex]

then... I really don't know what to do from there... do I just simply things?
Well, how about replacing that y in RHS with Cx2?
 
[tex]2Cx = \frac {2Cx^2}{x}[/tex]

[tex]2Cx = 2Cx[/tex]

is that possible?
 
Dr Game said:
[tex]2Cx = \frac {2Cx^2}{x}[/tex]

[tex]2Cx = 2Cx[/tex]

is that possible?

Yes, that's fine. :smile:
 

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