Method of Undetermined Coefficients

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SUMMARY

The discussion focuses on solving the differential equation y'' + y = sin(x) + xcos(x) using the Method of Undetermined Coefficients. The user initially derived a temporary particular solution yp as Asin(x) + Bcos(x) + (Cx + D)sin(x) + (Ex + F)cos(x) but encountered duplicates in the homogeneous solution. To address this, they multiplied the temporary solution by x, resulting in a more complex expression. Ultimately, the user resolved their confusion after several hours of analysis.

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  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with the Method of Undetermined Coefficients.
  • Knowledge of homogeneous and particular solutions in differential equations.
  • Ability to compute derivatives and simplify trigonometric expressions.
NEXT STEPS
  • Study the Method of Undetermined Coefficients in detail, focusing on examples with non-homogeneous terms.
  • Learn how to identify and handle duplicates in the homogeneous solution.
  • Practice solving various second-order linear differential equations with different non-homogeneous terms.
  • Explore alternative methods for solving differential equations, such as the Variation of Parameters.
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Students and educators in mathematics, particularly those studying differential equations, as well as anyone seeking to deepen their understanding of the Method of Undetermined Coefficients.

Chemmjr18
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Homework Statement


Find a particular solution yp of the given equation:
y''+y=sinx+xcosx

Homework Equations

The Attempt at a Solution


1) First I got the temporary yp: Asinx+Bcosx+(Cx+D)sinx+(Ex+F)cosx

2) Then I got the associated homogenous eq. check for to check for duplicates: C1cosx+C2sinx

3) Since there were duplicates, I multiplied the temporary solution by x: Axsinx+Bxcosx+(Cx2+Dx)sinx+(Ex2+Fx)cosx

4) I obtained the 2nd derivative of this and refined it: (2A+2C+2F)cosx+(-2B-2D+2E)sinx+(4E-D-B)xcosx+(-A-4C-F)xsinx-Cx2cosx-Ex2sinx

5) I then added this to 2) to get (2A+2C+2F)cosx+(-2B-2D+2E)sinx+(4E)xcosx+(-4C)xsinx=sinx+xcosx

I have more unknowns than I do equations, so I can't solve. Where'd I go wrong? I've been combing through the problem for almost 4 hours...
 
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Chemmjr18 said:

Homework Statement


Find a particular solution yp of the given equation:
y''+y=sinx+xcosx

Homework Equations

The Attempt at a Solution


1) First I got the temporary yp: Asinx+Bcosx+(Cx+D)sinx+(Ex+F)cosx

2) Then I got the associated homogenous eq. check for to check for duplicates: C1cosx+C2sinx

3) Since there were duplicates, I multiplied the temporary solution by x: Axsinx+Bxcosx+(Cx2+Dx)sinx+(Ex2+Fx)cosx

4) I obtained the 2nd derivative of this and refined it: (2A+2C+2F)cosx+(-2B-2D+2E)sinx+(4E-D-B)xcosx+(-A-4C-F)xsinx-Cx2cosx-Ex2sinx

5) I then added this to 2) to get (2A+2C+2F)cosx+(-2B-2D+2E)sinx+(4E)xcosx+(-4C)xsinx=sinx+xcosx

I have more unknowns than I do equations, so I can't solve. Where'd I go wrong? I've been combing through the problem for almost 4 hours...
Nevermind, I figured it out.
 

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