Solving a Simple Differential Equation: y''+y=5cos(x)

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SUMMARY

The discussion focuses on solving the differential equation y'' + y = 5cos(x). The general solution involves the homogeneous solution y = Acos(x) + Bsin(x), while the particular solution must be linearly independent from these terms. Participants suggest using the method of undetermined coefficients, specifically trying forms like y = xAcos(x) + xBsin(x) to find the particular solution. The importance of substituting guesses into the equation to determine coefficients is emphasized, along with the need for a systematic approach rather than trial and error.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with the method of undetermined coefficients
  • Knowledge of trigonometric functions and their derivatives
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the method of undetermined coefficients in detail
  • Learn about the Wronskian and its application in solving differential equations
  • Explore variations of parameters for finding particular solutions
  • Practice solving similar differential equations with different right-hand side functions
USEFUL FOR

Students studying differential equations, educators teaching advanced mathematics, and anyone looking to improve their problem-solving skills in applied mathematics.

physicsnewb7
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Homework Statement


y''+y=5cos(x)


Homework Equations



y=Acos(x)+Bsin(x)
e^ix=cos(x)+isin(x)

The Attempt at a Solution


I tried using compbinations of these equations to try and solve but I'm lost. Any suggestions would be greatly appreciated!
 
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y=Acos(x)+Bsin(x)

is the solution to the homogenous equation. The particular solution must be linearly independent from them. Try multiplying by some power of x to generate a guess for the particular solution.
 
try multiplying what by a power of x?
 
physicsnewb7 said:
try multiplying what by a power of x?

The solutions to the homogeneous equation. Try A xa cos x and B xb sin x and see if a and b can be chosen to get a solution.
 
is y= -2.5xcos(x) right?
 
Recheck your algebra, the sin term is actually the particular solution.
 
oh yeah i know i put cos by accident, but that is the solution right? I arrived at this solution by a long trial and error method. any way to quickly get at it in general?
 
physicsnewb7 said:
oh yeah i know i put cos by accident, but that is the solution right? I arrived at this solution by a long trial and error method. any way to quickly get at it in general?

You have a trig function on the RHS, so the particular solution must be related to a trig function. The sin and cos are already used in the solutions to the homogeneous equations, so the particular solution can't just be sin or cos. Hence you try powers of x times sin and cos. It's just a matter of substituting the guess into the equation and examining powers and coefficients. You could even guess that you should try x cos x and x sin x because there are no powers of x on the RHS.
 
Thank you very much you have been very helpful. I appreciate it
 

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