Working backwords from solution to DE

  • Thread starter Panphobia
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In summary, the conversation is discussing the solution of a second order differential equation with a non-homogeneous term. The poster has already determined the values of a, b, and c, and has identified the associated homogeneous DE. They are now seeking guidance on how to handle the term sintcost in order to find the complete solution. The solution is found by using the trigonometric identity \sin t \cos t = \frac12 \sin(2t). After receiving assistance, the poster thanks the responder for their help.
  • #1
Panphobia
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Homework Statement


if y = c1*e^t*sin(√3t)+c2*e^t*cos(√3t) + sintcost

The Attempt at a Solution



I have already figured out that (2±√-12)/2 so b = -2 and -12 = 4-4c so c = 4 and a = 1
and that means that the associated homogeneous DE is y'' -2y' + 4y = 0, but I don't know what to do about the sintcost to get then g(t) in y'' -2y' + 4y = g(t), I have tried looking at different methods of solving 2nd order differential equations and tried working backwards but I can't get anything going. What would a good first step be?
 
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  • #2
Panphobia said:

Homework Statement


if y = c1*e^t*sin(√3t)+c2*e^t*cos(√3t) + sintcost

The Attempt at a Solution



I have already figured out that (2±√-12)/2 so b = -2 and -12 = 4-4c so c = 4 and a = 1 and that means that the associated homogeneous DE is y'' -2y' + 4y = 0,but I don't know what to do about the sintcost to get then g(t) in y'' -2y' + 4y = g(t), I have tried looking at different methods of solving 2nd order differential equations and tried working backwards but I can't get anything going. What would a good first step be?

[itex]\sin t \cos t = \frac12 \sin(2t)[/itex].
 
  • #3
Yea I got the answer now, thanks a lot!
 

1. What is the process of working backwards from a solution to a differential equation?

When working backwards from a solution to a differential equation, the first step is to identify the known solution. Then, the solution is plugged into the original differential equation to find the unknown function or variable. This is typically done by using algebraic manipulation or integration techniques.

2. Why is working backwards from a solution to a differential equation useful?

This technique can be useful for solving complex differential equations that cannot be solved directly. It also allows us to check the validity of a given solution by verifying if it satisfies the original differential equation.

3. Is working backwards from a solution to a differential equation always possible?

No, it is not always possible. In some cases, the solution may not provide enough information to determine the original differential equation. Additionally, if the solution is not unique, it may not be possible to determine the original differential equation.

4. What are some common applications of working backwards from a solution to a differential equation?

This technique is commonly used in physics and engineering to model real-world systems. It can also be applied in economics, biology, and other scientific fields to understand and predict the behavior of various phenomena.

5. Are there any limitations to working backwards from a solution to a differential equation?

One limitation is that it can be a time-consuming and complex process, especially for higher-order differential equations. Additionally, it may not always be possible to find a closed-form solution, and numerical methods may be required instead.

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