General Solution to Non-homologous ODEs

Click For Summary
SUMMARY

The general solution to the non-homologous ordinary differential equation (ODE) y'' + y' + 4y = 2sinh(t) is given by y = C1*e^(-t/2)cos(√15t/2) + C2*e^(-t/2)sin(√15t/2) + (1/6)e^t - (1/4)e^(-t). The solution consists of a homogeneous part and a particular solution. The homogeneous solution is derived from the characteristic equation with roots r = [1 ± √(-15)]/2. The particular solution is obtained by assuming a form y_p = Ae^t + Be^(-t) and equating coefficients.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with the method of undetermined coefficients
  • Knowledge of characteristic equations and their roots
  • Basic understanding of hyperbolic functions, specifically sinh(t)
NEXT STEPS
  • Study the method of undetermined coefficients in depth
  • Learn how to solve homogeneous and non-homogeneous ODEs
  • Explore the characteristics of hyperbolic functions and their applications
  • Practice solving ODEs with varying coefficients and different forcing functions
USEFUL FOR

Students studying differential equations, mathematics educators, and anyone seeking to enhance their understanding of non-homogeneous ODE solutions.

zabumafu
Messages
13
Reaction score
0

Homework Statement


Find the general solution of the given differential equation:

y''+y'+4y=2sinht


Homework Equations



I believe sinht=(e^t-e^-t)/2

The Attempt at a Solution



I tried to find the general equation if it were homogenous however I get the roots are
r=[1+- (-15)^.5]/2 and get stuck. If anyone can help me figure out the rest of the problem I should be able to teach myself how to do the rest of them. I know the answer is:

y1=C1*e^(-t/2)cos(root(15t/2))+C2*e^(-t/2)sin(root(15t/2))+1/6e^t-1/4e^-t
 
Physics news on Phys.org
You see how the final answer looks like and still you don't know how to answer this?

In the final answer the first two terms are the solution of the homogenous DE, and the rest two terms are the private solution, i.e you guess: y_p = Ae^t+Be^-t and then plug it to the DE, and equate the coeffiecients on both sides of the equation, such that the coeff of e^t on one side is the same on the other side, the same with e^-t, this is how you find A and B.
 

Similar threads

Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
2K