2nd order non homogeneous equation

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SUMMARY

The discussion centers on solving the second-order non-homogeneous differential equation y'' + y = -2 Sin(x). The homogeneous solution is correctly identified as yh(x) = C1 Cos(x) + C2 Sin(x). The user struggles to find the particular solution, initially proposing y = A Cos(x) + B Sin(x), which fails due to linear dependence with the homogeneous solution. The correct approach involves using a particular solution of the form y = x(A Cos(x) + B Sin(x)) to ensure linear independence.

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  • Understanding of second-order differential equations
  • Familiarity with homogeneous and particular solutions
  • Knowledge of linear independence in function spaces
  • Basic skills in trigonometric identities and calculus
NEXT STEPS
  • Study the method of undetermined coefficients for non-homogeneous equations
  • Learn about the Wronskian and its role in determining linear independence
  • Explore the use of variation of parameters for finding particular solutions
  • Practice solving similar second-order differential equations with different non-homogeneous terms
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Students studying differential equations, mathematics educators, and anyone seeking to deepen their understanding of non-homogeneous differential equations and their solutions.

mitch987
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Homework Statement


y'' + y = -2 Sinx


Homework Equations





The Attempt at a Solution


finding the homogeneous solution, is simple;
yh(x) = C1 Cos(x) + C2 Sin(x)

for the particular solution,
I let y = A Cos(x) + B Sin(x)
thus, y' = -A Sin(x) + B Cos(x)
y'' = -A Cos(x) - B Sin(x)

substituting these into the differential equation;
-A Cos(x) - B Sin(x) + A Cos(x) + B Sin(x) = -2 Sin(x)
0 = -2 Sin(x)

This is where i get stuck as I'm unable to find any values for A & B and hence, cannot find the particular solution in order to find the general solution y(x) = yh(x) + yp(x)
Any help would be greatly appreciated.
Thanks,
 
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The particular solution must be linearly independent of the homogeneous solutions. Since those are already [tex]\sin x[/tex] and [tex]\cos x[/tex], you should try [tex]x\sin x[/tex] and [tex]x\cos x[/tex].
 
of course! Forgot that the particular solution cannot equal the homogeneous solution.
Thanks for your help.
 

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