Approximating k with Shooting Method and Euler's Method

  • Context: Undergrad 
  • Thread starter Thread starter androol
  • Start date Start date
  • Tags Tags
    Method Shooting method
Click For Summary
SUMMARY

The discussion focuses on approximating the constant k in the differential equation dy/dt = ky using the shooting method and Euler's method. The initial conditions are y(0) = 1 and y(10) = 4. The procedure involves estimating k by taking a time step of dt = 5, leading to the quadratic equation 25k² + 10k - 3 = 0, which can be solved for k. After obtaining an initial estimate of k, users are advised to refine their approximation by increasing the number of steps in Euler's method until the desired outcome is achieved.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with the shooting method for boundary value problems.
  • Knowledge of Euler's method for numerical integration.
  • Basic algebra skills for solving quadratic equations.
NEXT STEPS
  • Learn about the shooting method in detail, including its advantages and limitations.
  • Explore advanced numerical methods for solving differential equations, such as Runge-Kutta methods.
  • Study the impact of varying step sizes in Euler's method on accuracy and convergence.
  • Investigate the application of numerical methods in real-world scenarios, such as population modeling or chemical reactions.
USEFUL FOR

Mathematicians, engineers, and students involved in numerical analysis, particularly those working with differential equations and numerical methods for approximating solutions.

androol
Messages
9
Reaction score
0
dy/dt = ky, where k is a constant.
y|t=0 = 1; y|t=10 = 4

I need to approximate k using the shooting method and Euler's method... please help.. :)

wat's the procedure to approximate k using the shooting method :confused:
 
Physics news on Phys.org
Do you know what "Euler's method" is?

Start with an easy estimate: take dt= 5 so we have only two steps
Initially, t= 0, y= 1, so dy/dt= k and dy= 5k

then t= 5, y= 1+ dy= 1+ 5k so dy/dt= k(1+ 5k)= k+ 5k2 and so dy= 5(k+ 5k2)= 5k+ 25k2.

Finally, then, x= 10, y= 1+ 5k+ dy= 1+ 5k+ 5k+ 25k2= 1+ 10k+ 26k2= That is, 25k2+ 10k- 3= 0. Solve that for k.

Now that you know, roughly, k, use, say, 10 steps from x= 0 to x= 10 with that k and see if you get 4 there. You won't of course, but that will tell you whether to make k smaller or larger. Do that repeatedly until you get a good method.
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K