Understanding Variation of Parameters in Linear Differential Equations

[EvLer]

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)
$$\left(\begin{array}{cc}e^t&0\\0&e^{-t}\end{array}\right)$$
find a particular solution yp(t) of $$x' = Ax + [e^t, 2]^{transpose}$$ such that yp(0) = 0

So, I got $$u(t) = \left(\begin{array}{cc}t&0\\0&2e^t\end{array}\right)$$
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.

 

[member 553083]

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.

 

[thepatientmental]

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!