Third order Euler Differential Equation

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Homework Help Overview

The problem involves a third order Euler differential equation given by the expression x^3y'''-3x^2y''+7xy'-8y=x+e^2x. Participants are exploring potential solution methods and discussing the implications of various approaches.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants suggest trying a solution of the form y=x^{n} to simplify the problem, while others express uncertainty about how this approach would work given specific instructions to use t=lnx. There is discussion about the resulting cubic equation and its coefficients.

Discussion Status

Participants are actively engaging with different methods and sharing calculations related to the cubic equation derived from the proposed solution form. There is a recognition of potential solutions, but no consensus has been reached on the best approach.

Contextual Notes

Some participants mention constraints such as being instructed to use t=lnx, which may affect their approach to solving the differential equation.

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Homework Statement


x^3y'''-3x^2y''+7xy'-8y=x+e^2x

Edit : Got it thanks
 
Last edited:
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Have you thought about trying the solution y=x^{n}? This might lead to an easier cubic to solve.
 
hunt_mat said:
Have you thought about trying the solution y=x^{n}? This might lead to an easier cubic to solve.

Not sure how that would work out, only told to use t=lnx. But using y=x^{n} won't cancel the coefficients to be constants.

Edit: I looked up your response on Wikipedia and found something similar on it, never learned to do it this way tho. I'll give it a shot.
 
Last edited:
They will add up to give x^{n} though. for example if y=x^n, then x^{3}y'''(x)=n(n-1)(n-2)x^n, likewise for x^2 and so on. I calculate that the cubic you arrive at the cubic:
<br /> n^{3}-6n^{2}+12n-8<br />
One of the solutions look to be n=2.

Mat
 
hunt_mat said:
They will add up to give x^{n} though. for example if y=x^n, then x^{3}y'''(x)=n(n-1)(n-2)x^n, likewise for x^2 and so on. I calculate that the cubic you arrive at the cubic:
<br /> n^{3}-6n^{2}+12n-8<br />
One of the solutions look to be n=2.

Mat

Got it thanks.
 

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