Assistance with Third Order Differential Equation

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Homework Help Overview

The discussion revolves around solving a third-order differential equation of the form y''' + 8y = xsin(2x). Participants are exploring the methods to approach this problem, particularly focusing on the homogeneous part of the equation and the form of the particular solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to solve the homogeneous differential equation y''' + 8y = 0 and suggests a possible form for the particular solution involving trigonometric functions. Other participants question the complementary solution and its relation to the non-homogeneous term.

Discussion Status

Participants are actively engaging with each other's attempts, with some providing feedback on the proposed solutions and suggesting alternative representations. There is a focus on ensuring that the complementary solution is correctly identified before proceeding with the particular solution.

Contextual Notes

There appears to be some uncertainty regarding the appropriate form of the particular solution and its relationship to the homogeneous solution. Participants are also considering the implications of using complex exponentials versus trigonometric functions.

kelvin2013
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Homework Statement


Hi,

Just wondering if anyone knows how to solve the following as I am not sure where to start at all:

y''' + 8y = xsin(2x)

Any help would be great.


Homework Equations






The Attempt at a Solution



I'm thinking solving the homogeneous DE y'''+8y = 0 and then as far as I can see a possible solution maybe of the form yp=Axsin(2x)+Bxcos(2x)+Csin(2x)+Dcos(2x) ?
 
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kelvin2013 said:

Homework Statement


Hi,

Just wondering if anyone knows how to solve the following as I am not sure where to start at all:

y''' + 8y = xsin(2x)

Any help would be great.


Homework Equations






The Attempt at a Solution



I'm thinking solving the homogeneous DE y'''+8y = 0 and then as far as I can see a possible solution maybe of the form yp=Axsin(2x)+Bxcos(2x)+Csin(2x)+Dcos(2x) ?

What do you get for ##y_c## when you solve the homogeneous DE? You need to know that before you can predict the form of the NH equation.
 
Hi,

thanks for the reply - I get:

yp=c1e-2x+c2e(1+i√3)x+c3e(1-i√3)x

?
 
kelvin2013 said:
Hi,

thanks for the reply - I get:

yp=c1e-2x+c2e(1+i√3)x+c3e(1-i√3)x

?

OK. So far so good. It would be nicer to use sines and cosines instead of the complex exponentials. Anyway, your complementary solution doesn't contain any terms like the NH term. So what happens when you try your proposed ##y_p##?
 

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