Help with initial value problem(IVP)

  • Context: Graduate 
  • Thread starter Thread starter shayaan_musta
  • Start date Start date
  • Tags Tags
    Initial Value
Click For Summary
SUMMARY

The discussion focuses on the correct application of general solutions for ordinary differential equations (ODEs) based on the nature of the roots derived from the characteristic equation. The three general solutions are: 1) \(y(t) = c_1 e^{m_1 t} + c_2 e^{m_2 t}\) for distinct real roots (D > 0), 2) \(y(t) = c_1 e^{mt} + c_2 t e^{mt}\) for repeated real roots (D = 0), and 3) \(y(t) = c_1 e^{u \cos(v)} + c_2 \sin(v)\) for complex roots (D < 0). The specific example discussed is the equation \(y'' + 6y' + 9y = 0\), which has a discriminant of 0, indicating repeated roots, thus necessitating the use of the second solution.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with characteristic equations and their roots
  • Knowledge of discriminants in quadratic equations
  • Basic concepts of exponential functions and their applications in ODEs
NEXT STEPS
  • Study the derivation and application of the characteristic equation for ODEs
  • Learn about the implications of the discriminant in determining the nature of roots
  • Explore the method of undetermined coefficients for solving ODEs
  • Investigate the use of Laplace transforms in solving linear differential equations
USEFUL FOR

Students and professionals in mathematics, engineering, and physics who are dealing with ordinary differential equations and require a clear understanding of solution methods based on the nature of roots.

shayaan_musta
Messages
208
Reaction score
2
Hello experts!

As we know that there are 3 different general solutions of an ordinary differential equation depends on that what type of roots we've gotten in the solution, listed below.
1) y(t)=c1em1t+c2em2t
2) y(t)=c1emt+c2temt
3) y(t)=c1eucos(v)+c2sin(v)


Now 1st solution is used when discriminant i.e. D>0, 2nd is used when D=0 and 3rd is used when D<0

But here is a question,
y''+6y'+9y=0
By solving we get roots,
m1=-3 and m2=-3
While discriminant is -30
& book has used 2nd solution to solve it, if D<0 it must use 3rd solution.

How do we choose correct solution? Does solution is according to nature of roots(discriminant) or m1 & m2?

Thanks for your contribution.
 
Physics news on Phys.org
Discriminant is not -30, it is 6^2-4.1.9=0
 
Ok.
Thanks to light me the right way.
 

Similar threads

Graduate Volterra equation of the second kind
  • · Replies 1 ·
Replies
1
Views
963
MHB -2.4.2 interval of initial value problem
  • · Replies 4 ·
Replies
4
Views
2K
MHB Solving an Initial Value Problem Analytically
  • · Replies 5 ·
Replies
5
Views
2K
MHB 3.1.11 find the solution of the given initial value problem:
  • · Replies 4 ·
Replies
4
Views
2K
MHB -1.8.17 initial value then find C
  • · Replies 11 ·
Replies
11
Views
2K
Undergrad Verifying properties of Green's function
  • · Replies 1 ·
Replies
1
Views
3K
MHB -b.2.2.26 IVP min value y'=2(1+x)(1+y^2); y(0)=0
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Correct Upper/Lower limits for continuation of solutions for ODE
  • · Replies 0 ·
Replies
0
Views
4K
MHB De1.2.1 Solve the following initial value problem
  • · Replies 7 ·
Replies
7
Views
3K
Undergrad Understanding Euler Method: Finding Initial Condition of y(0)=1
  • · Replies 7 ·
Replies
7
Views
5K