Second order nonhomogeneous ODE

Click For Summary
SUMMARY

The discussion focuses on solving the second order nonhomogeneous ordinary differential equation (ODE) given by y'' + 3y' + 3.25 = 3cos(t) - 1.5sin(t). The homogeneous solution is correctly identified as e^(-1.5t)(Acos(t) + Bsin(t)). The participant initially struggled with the particular solution but realized that using K1cos(ωt) + K2cos(ωt) is redundant due to linear combinations, leading to a simplified approach that yielded the correct answer.

PREREQUISITES
  • Understanding of second order ordinary differential equations (ODEs)
  • Familiarity with homogeneous and particular solutions
  • Knowledge of trigonometric functions and their linear combinations
  • Experience with solving differential equations using the method of undetermined coefficients
NEXT STEPS
  • Study the method of undetermined coefficients for nonhomogeneous ODEs
  • Learn about the Wronskian and its application in solving ODEs
  • Explore the concept of linear independence in the context of differential equations
  • Investigate the use of Laplace transforms for solving ODEs
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as engineers and physicists applying ODEs in their fields.

yaro99
Messages
75
Reaction score
0

Homework Statement



y''+3y'+3.25=3cost-1.5sint

Homework Equations


yh = e(a/2)t(Acost+Bsint)
yp = Kcos(ωt)+Msin(ωt) [when r(x)=kcos(ωt) or ksin(ωt)]

The Attempt at a Solution



I got the homogeneous solution, which is e-1.5t(Acost+Bsint)
but I am having trouble with the particular solution.

I tried the above equation, making yp=K1cos(ωt)+M1sin(ωt)+K2cos(ωt)+M2sin(ωt)
since there are 2 trig functions as r(t).
I couldn't solve for the variables by plugging into the original equation because I was left with 4 variables and only 2 equations.

EDIT: realized I wasn't consistent with my independent variable, made them all t's instead of t's and x's
 
Last edited:
Physics news on Phys.org
K1cos(ωx) + K2cos(ωx) = (K1 + K2)cos(ωx). Having two sin/cos functions is redundant, since they are linear combinations (in this case, they are the same function, entirely).
 
djh101 said:
K1cos(ωx) + K2cos(ωx) = (K1 + K2)cos(ωx). Having two sin/cos functions is redundant, since they are linear combinations (in this case, they are the same function, entirely).

Ah I get it now. I solved for the values (M1+M2) and (K1+K2) instead of the variables individually and got the right answer. Thanks!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K