Understanding Variation of Parameters in Linear Differential Equations

In summary, the conversation is about finding a particular solution to the system x' = Ax + [e^t, 2]^{transpose}. The solution is given by yp(t) = X(t)*u(t) + \int_0^t X(s) ds * [e^t,2]^T, where X(t) is the fundamental matrix of the system and u(t) is a specific function. The person asking the question was confused about the integration of u'(t), but the solution was clarified by including the integration in the particular solution.
  • #1
EvLer
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I have been trying to get this for a while and can't figure it out:

if a fundamental matrix of the system x' = Ax is X(t)
[tex]\left(\begin{array}{cc}e^t&0\\0&e^{-t}\end{array}\right)[/tex]
find a particular solution yp(t) of [tex] x' = Ax + [e^t, 2]^{transpose}[/tex] such that yp(0) = 0

So, I got [tex]u(t) = \left(\begin{array}{cc}t&0\\0&2e^t\end{array}\right)[/tex]
but when I multiply X(t) by it, I do not get the right answer. Is there a coefficient involved with integration of u'(t)?

Thanks for help.
 
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  • #2
Yes, you need to integrate u'(t) in order to find the particular solution yp(t). The particular solution is given by:yp(t) = X(t)*u(t) + \int_0^t X(s) ds * [e^t,2]^Twhere X(s) = \left(\begin{array}{cc}e^s&0\\0&e^{-s}\end{array}\right)and \int_0^t X(s) ds = \left(\begin{array}{cc}1-e^t&0\\0&e^{-t}-1\end{array}\right). Thus, yp(t) = \left(\begin{array}{cc}te^t-e^{2t}+1&0\\0&2e^{2t}-2e^t-1\end{array}\right). It can be easily checked that yp(0) = 0.
 
  • #3


Variation of parameters is a method used to find a particular solution to a non-homogeneous linear differential equation. In this case, we are given a fundamental matrix X(t) for the system x' = Ax, and we are trying to find a particular solution yp(t) for the system x' = Ax + [e^t, 2]^{transpose}.

To find yp(t), we can use the formula yp(t) = X(t)∫X^-1(t)b(t)dt, where b(t) is the non-homogeneous term [e^t, 2]^{transpose}.

In this case, X(t) = \left(\begin{array}{cc}e^t&0\\0&e^{-t}\end{array}\right) and X^-1(t) = \left(\begin{array}{cc}e^{-t}&0\\0&e^t\end{array}\right).

Therefore, we have yp(t) = \left(\begin{array}{cc}e^t&0\\0&e^{-t}\end{array}\right)∫\left(\begin{array}{cc}e^{-t}&0\\0&e^t\end{array}\right)\left(\begin{array}{c}e^t\\2\end{array}\right)dt = \left(\begin{array}{c}e^t\\2e^t\end{array}\right).

This particular solution satisfies the given system and also yp(0) = 0, as required.

In your attempt, you have correctly found u(t) = \left(\begin{array}{cc}t&0\\0&2e^t\end{array}\right), but you have not integrated u'(t) correctly. The correct integration would be u(t) = \left(\begin{array}{c}t\\2e^t\end{array}\right) + c\left(\begin{array}{c}e^t\\0\end{array}\right), where c is a constant of integration.

When you multiply X(t) by this u(t), you should get the correct particular solution. I hope this helps clarify the concept of variation of parameters for you. If you are still having trouble, I suggest seeking further guidance from your instructor or a tutor. Good luck!
 

FAQ: Understanding Variation of Parameters in Linear Differential Equations

1. What is the concept of variation of parameters in mathematics?

The variation of parameters is a mathematical technique used to solve non-homogeneous linear differential equations. It involves finding a particular solution by assuming that the coefficients of the general solution are functions of the independent variable.

2. When is the variation of parameters method used?

The variation of parameters method is used when solving non-homogeneous linear differential equations with constant coefficients. It is an alternative method to the method of undetermined coefficients, which only works for certain types of non-homogeneous equations.

3. How does the variation of parameters method differ from other methods of solving differential equations?

The variation of parameters method differs from other methods, such as the method of undetermined coefficients, because it allows for a wider range of non-homogeneous equations to be solved. It also provides a more general solution by including arbitrary constants in the general solution.

4. What is the process for using the variation of parameters method to solve a differential equation?

The process for using the variation of parameters method involves the following steps:

  • Find the general solution of the associated homogeneous equation.
  • Assume that the coefficients of the general solution are functions of the independent variable.
  • Substitute these functions into the non-homogeneous equation and solve for the functions.
  • Add the particular solution to the general solution to get the complete solution.

5. Are there any limitations or drawbacks to using the variation of parameters method?

One limitation of the variation of parameters method is that it can be more time-consuming and complicated compared to other methods. It also may not work for all types of non-homogeneous equations, in which case other methods, such as Laplace transforms, may be more suitable. Additionally, the resulting solution may involve more arbitrary constants, which can make it less intuitive to interpret.

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