Differential Equation NonHomogeneous EQ

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SUMMARY

The discussion focuses on solving the non-homogeneous differential equation y'' + 4y = sin²(x). The correct particular solution is given as yp(x) = (1/8)(1 - x sin(2x)), which differs from the user's derived solution of yp = 1/2 - (1/2)x sin(2x). The discrepancy arises from the user's incorrect transformation of sin²(x) to 4sin²(x) instead of using the double angle identity sin²(x) = 1/2 - (1/2)cos(2x). The homogeneous solution is correctly identified as yh = C1cos(2x) + C2sin(2x).

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with the method of undetermined coefficients for finding particular solutions.
  • Knowledge of trigonometric identities, particularly the double angle identity for sine.
  • Ability to perform algebraic manipulations involving derivatives and substitutions.
NEXT STEPS
  • Review the method of undetermined coefficients in solving non-homogeneous differential equations.
  • Study the double angle identities in trigonometry, focusing on sin²(x) and its transformations.
  • Practice solving second-order linear differential equations with various non-homogeneous terms.
  • Explore the implications of homogeneous solutions in the context of differential equations.
USEFUL FOR

Students studying differential equations, educators teaching advanced mathematics, and anyone seeking to deepen their understanding of non-homogeneous differential equations and their solutions.

JJBrian
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Homework Statement



Find a particular solution of the given differential equation.
y'' + 4y = (sin^2)(x)

Answer from book: yp(x)= (1/8)(1-xsin2x)


The Attempt at a Solution



y'' + 4y = 4(sinx)^2....(1)

4(sinx)^2=2 - 2cos2x

Homogeneous solution:
r^2+4=0
r^2= -4
r=+-2i ==>
yh=C1cos(2x)+C2sin(2x)

Particular solution:
yp=A+Bxcos2x+Cxsin2x
yp' =Bcos2x-2Bxsin2x+Csin2x+2Cxcos2x

yp" = -2Bsin2x -2Bsin2x-4Bxcos2x
+2Ccos2x+2Ccos2x - 4Cxsin2x

Substitute y and y" in (1) ==>
4A - 4Bsin2x+4Ccos2x=2 - 2cos2x ==>

A=1/2
B=0
C= -1/2

yp=1/2 - (1/2)xsin2x

This is answer is different from the book. What am I doing wrong?
 
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JJBrian said:

Homework Statement



Find a particular solution of the given differential equation.
y'' + 4y = (sin^2)(x)

Answer from book: yp(x)= (1/8)(1-xsin2x)


The Attempt at a Solution



y'' + 4y = 4(sinx)^2....(1)
How did the right side go from sin2(x) to 4sin2(x)?

What you need to do is rewrite the right side using a double angle identity, sin2(x) = 1/2 - (1/2)cos(2x).
Your work below suggests that you have done this, but I don't see that you mention it anywhere.
JJBrian said:
4(sinx)^2=2 - 2cos2x

Homogeneous solution:
r^2+4=0
r^2= -4
r=+-2i ==>
yh=C1cos(2x)+C2sin(2x)
This is fine.
JJBrian said:
Particular solution:
yp=A+Bxcos2x+Cxsin2x
This (above) is what you need.
JJBrian said:
yp' =Bcos2x-2Bxsin2x+Csin2x+2Cxcos2x

yp" = -2Bsin2x -2Bsin2x-4Bxcos2x
+2Ccos2x+2Ccos2x - 4Cxsin2x

Substitute y and y" in (1) ==>
4A - 4Bsin2x+4Ccos2x=2 - 2cos2x ==>

A=1/2
B=0
C= -1/2

yp=1/2 - (1/2)xsin2x

This is answer is different from the book. What am I doing wrong?
The only mistake I see is that you changed the right side of the DE from sin2(x) to 4sin2(x). That could be causing problems for you in solving for the coefficients A, B, and C.
 

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