# Initial-value problem

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Using the method of undetermined coefficients, find the solution of the system:

X'=AX + B

that satisfies the initial condition:

X(0)=( 0

1

-1).

A and B are matrices defined in the attached Notepad file.

Note: When solving the homogeneous soln, exhibit a fundamental matrix psi(t) and also the special fundamental matrix phi(t) satisfying phi(0)=I.

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##### Solution Summary

This shows how to find the solution of a system using the method of undetermined coefficients.

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Solution:For this system X'=AX+B (1)

We first need to know the given matrices A and B. If they are matrices with constant numbers, it is easy to solve them. For example, if we know A=P'CP where C=diag(a1,a2,...,an) and P'=P^(-1). P^(-1) is the inverse of P. Note that such P is called orthogonal matrix. Many matrices can easily find such matrix P such that A=P'CP.

Then substituting A=P^(-1)CP , (1) becomes

PX'=CPX+PB (2)

where C=diag{a1,a2,...,an}.

Let Y=PX and D=PB,then (2) is as follows

Y'=CY+D (3)

Obviously Y=-C^(-1)D is a special solution for (3). In order to solve (3), we only need to solve the linear diff. equations (4) as follows

Y'=CY (4)

In fact, IF Y=(y1,y2,...,yn), then (4) can rewrite as follows.

...

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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