Finding the particular solution to an ODE with set boundary conditions.

In summary, the conversation discusses a problem with a boundary condition of dx/dt=-2 and t=0. The initial answer to the auxiliary equation is incorrect, and the correct homogeneous solution is x_{h}=C_{1}cos(2t)+C_{2}Sin(2t). The completed solution for part c) is x=(i/sqrt3)[-e^(2it)+e^(-2it)] and y_{h}=C_{1}sin2t+C_{2}cos2t, with C_{1}=0 and C_{2}=1. The use of Euler's formula in this solution is questioned, and the conversation concludes by suggesting a resource for further explanation.
  • #1
Jack_O
65
0

Homework Statement



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Homework Equations



N/A

The Attempt at a Solution



5184b1eb.jpg


The problem and attempt are as above, I'm not sure where to go from here though. I'm not sure what to do with the boundary condition of dx/dt=-2 and t=0.
Any help appreciated.
 
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  • #2
The answer to your auxiliary equation is wrong, m=+/- 2i not +/- 2 so the homogeneous solution is of the form [tex]x_{h}=C_{1}cos(2t)+C_{2}Sin(2t)[/tex]
 
  • #3
Thanks for pointing that out, I've had another go and realized i needed to differentiate the general solution and then sub in the other boundaries to get the other simultaneous equation. My completed solution looks like this:

742403f1.jpg


I have also done part c), got the answer x=(i/sqrt3)[-e^(2it)+e^(-2it)]
 
  • #4
make your solution without complex coefficients so [tex]y_{h}=C_{1}sin2t+C_{2}cos2t[/tex]

then work out the result [tex]C_{1}sin0+C_{2}cos0=1[/tex]
[tex]C_{2}=1[/tex]

[tex]y_{h}'=2C_{1}cos2t-2C_{2}sin2t[/tex]

[tex]2C_{1}cos0-2C_{2}0=-2[/tex]

[tex]-2C_{1}-2=-2[/tex]

[tex]C_{1}=0[/tex]...
 
  • #5
If i use Euler's formula to sub out e^(2it) for cos(2t)+i*sin(2t) won't i still be left with complex coefficients? Their are complex numbers in the (1+i) term. I don't see how i can cancel out all the complex numbers.
 
  • #6

1. What is a particular solution to an ODE?

A particular solution to an ODE (ordinary differential equation) is a specific function that satisfies the given equation and its initial or boundary conditions. It is a unique solution that fits the given parameters and is different from the general solution, which contains arbitrary constants.

2. How do you find the particular solution to an ODE?

The particular solution to an ODE can be found by substituting the given initial or boundary conditions into the general solution and solving for the arbitrary constants. Alternatively, it can be found using various mathematical techniques such as separation of variables, integrating factors, or variation of parameters.

3. What are boundary conditions in an ODE?

Boundary conditions in an ODE specify the values of the dependent variable (usually denoted by y) at specific points or boundaries of the independent variable (usually denoted by x). They are necessary to find a unique particular solution to the equation.

4. Can there be multiple particular solutions to an ODE with set boundary conditions?

No, there can only be one particular solution to an ODE with set boundary conditions. This is because the given conditions uniquely determine the solution and any other function that satisfies the equation but not the conditions is not considered as a particular solution.

5. What is the significance of finding the particular solution to an ODE?

Finding the particular solution to an ODE is important because it helps us to obtain a specific solution that fits the given conditions and can be used to model real-world phenomena. It also helps us to understand the behavior of the system described by the equation and make predictions about its future behavior.

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