Ordinary Differential Equations

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Homework Help Overview

The discussion revolves around ordinary differential equations, specifically examining the relationship between a given equation and a proposed solution involving constants.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants are exploring whether the proposed solution, involving a square root and a constant, satisfies the original differential equation. There is uncertainty regarding the role of the constant C1 in the context of the equations presented.

Discussion Status

The conversation is ongoing, with participants attempting to clarify the relationship between the equations and the proposed solution. Some guidance has been offered regarding finding derivatives and substituting them back into the original equation.

Contextual Notes

There is mention of confusion regarding the constants and their implications in the equations, as well as a potential misinterpretation of the second equation.

aaronfue
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Homework Statement



y = 2xy' + y(y')2; y2 = C1 (x + C1 /4)



2. The attempt at a solution

I thought that I could take the first equation and set it equal to zero. But the C1 in the second equation is throwing me off.

Am I supposed to set the second equal to C1?
 
Last edited:
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Are you saying that you want to show that [itex]y= \pm\sqrt{C_1(x+ \frac{C_1}{4})}[/itex] satisfies [/itex]y= 2xy'+ y(y')^2[/itex]?

Just find the derivative of y and plug it into the equation.
 
HallsofIvy said:
Are you saying that you want to show that [itex]y= \pm\sqrt{C_1(x+ \frac{C_1}{4})}[/itex] satisfies [/itex]y= 2xy'+ y(y')^2[/itex]?

Just find the derivative of y and plug it into the equation.

Sorry the second equation is y2, but I think I see what you are saying.
 
Yes, that's why I took the square root to get y.
 

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