How Does Euler's Method Work with Non-Zero Starting Points?

  • Context: Undergrad 
  • Thread starter Thread starter Ready2GoXtr
  • Start date Start date
  • Tags Tags
    Method
Click For Summary
SUMMARY

Euler's Method can be applied to non-zero starting points effectively. In the example provided, with the differential equation x' = 2x² + tx and the initial condition x(2) = 1, the correct calculation for x(2.25) is x(2.25) = x(2) + f(tn, xn) * h, where h = 0.25. Specifically, the calculation should use the value of t at the current step, leading to x(2.25) = 1 + f(2, 1) * 0.25, resulting in the new value based on the slope at that point.

PREREQUISITES
  • Understanding of differential equations and initial value problems
  • Familiarity with Euler's Method and its formula
  • Basic knowledge of calculus, particularly derivatives
  • Ability to perform numerical calculations and iterations
NEXT STEPS
  • Study the derivation and applications of Euler's Method in numerical analysis
  • Learn about improved methods such as Runge-Kutta for solving differential equations
  • Explore the concept of step size (h) and its impact on numerical solutions
  • Investigate the stability and error analysis of numerical methods for differential equations
USEFUL FOR

Mathematicians, engineering students, and anyone involved in numerical methods for solving differential equations will benefit from this discussion.

Ready2GoXtr
Messages
74
Reaction score
0
Every single example i have seen of eulers method starts with a range of 0 to some number, what if you had it going from like 2 to some number?

for example

if we had
x'=2x^2 + tx x(2)=1
and our range was
2 <= t <= 2.5 with h = .25

would I do this?

x(2.25)=x(2)+f(x(2),0)*.25 = 1.5

or would i do this

x(2.25)=x(2)+f(x(2),2)*.25 = 2
 
Physics news on Phys.org
Ready2GoXtr said:
Every single example i have seen of eulers method starts with a range of 0 to some number, what if you had it going from like 2 to some number?

for example

if we had
x'=2x^2 + tx x(2)=1
and our range was
2 <= t <= 2.5 with h = .25

would I do this?

x(2.25)=x(2)+f(x(2),0)*.25 = 1.5

or would i do this

x(2.25)=x(2)+f(x(2),2)*.25 = 2

You do the latter one. The idea is that at each step (tn,xn) have the point approximately on the curve and the slope at that point x' = f(tn,xn). So the new value xn+1 = xn + hf(tn,xn) where h is the step size.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
6K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K