Solving System of Coupled DEs: Parametrized Curve Solution

  • Context: Graduate 
  • Thread starter Thread starter Si-duck
  • Start date Start date
  • Tags Tags
    Coupled Curve System
Click For Summary
SUMMARY

The discussion focuses on solving a system of coupled differential equations defined by x' = 5x - y with initial condition x(0) = 6 and y' = -x + 5y with y(0) = -4. The parametrized curve solution is given as r(t) = (exp(4t) + 5exp(6t), exp(4t) - 5exp(6t)). To verify this solution, participants suggest differentiating the parametrized expressions for x(t) and y(t), substituting them back into the original differential equations, and confirming that both sides of the equations are equal.

PREREQUISITES
  • Understanding of coupled differential equations
  • Knowledge of differentiation and initial value problems
  • Familiarity with exponential functions and their properties
  • Ability to manipulate and substitute expressions in mathematical equations
NEXT STEPS
  • Study the method of solving coupled differential equations
  • Learn about initial value problems in differential equations
  • Explore the properties of exponential functions in differential equations
  • Practice verifying solutions by substitution in differential equations
USEFUL FOR

Mathematicians, engineering students, and anyone interested in solving and verifying solutions to systems of coupled differential equations.

Si-duck
Messages
1
Reaction score
0
Consider a system of coupled differential equations

x'=5x-y where x(0) = 6
y'=-x+5y where y(0)=-4

a) Show that the parametrised curve (x,y)= r(t)=(exp(4t) + 5exp(6t), exp(4t) - 5exp(6t))

How would you go about showing this?
 
Physics news on Phys.org


Si-duck said:
Consider a system of coupled differential equations

x'=5x-y where x(0) = 6
y'=-x+5y where y(0)=-4

a) Show that the parametrised curve (x,y)= r(t)=(exp(4t) + 5exp(6t), exp(4t) - 5exp(6t))

How would you go about showing this?

You just substitute them into the DEs. Keep in mind that x' is really dx/dt and y' is dy/dt. Ok then, differentiate the solutions, put them on the left sides, then substitute the solutions for x(t) and y(t) on the right and see if they're equal.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K