Mathematics, 26.02.2020 22:56 pandamaknae2003

Exercise 3.3.2: Verify that the system x® 0 1 3 3 1 x® has the two solutions 1 1 e 4t and 1 −1 e −2t a) . b) Write down the general solution. Write down the general solution in the form x1 ?, x2 ? (i. e. write down a formula for each element of the solution).

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Answer from: greystokey

The general solution of the system is

$X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\$

Step-by-step explanation:

To verify that the system $X^{'} =\left[\begin{array}{cc}1&3\\3&1\end{array}\right] X$ has the solutions given,

First, we determine the Eigen Values β of the matrix of the form $X^{'} =A X$ Using |A-βI|=0, where I is the Identity Matrix, we have:

$|A-\beta I|=0$

$|\left(\begin{array}{cc}1&3\\3&1\end{array}\right )-\left(\begin{array}{cc}\beta &0\\0&\beta \end{array}\right )|=0$

Subtracting matrices

$\left|\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right |=0$

Taking the determinant

$(1-\beta)(1-\beta)-(3X3)=0\\1-\beta-\beta+\beta^{2}-9=0\\\beta^{2}-2\beta-8=0\\\beta^{2}-4\beta+2\beta-8=0\\\beta(\beta-4)+2(\beta-4)=0\\(\beta-4)((\beta+2)=0\\\beta=4 or -2$

We determine the eigen vector using $\left(\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$

When $\beta$ = -2,

$\left(\begin{array}{cc}3 &3\\3&3 \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$ which implies that $3c_{11}+3c_{12}=0$ and thus

$If c_{11}=1, c_{12}=-1$

When $\beta$ = 4,

$\left(\begin{array}{cc}-3 &3\\3&-3 \end{array}\right)\left(\begin{array}{c}c_{21} &c_{22} \end{array}\right)=0$ which implies that $3c_{21}-3c_{22}=0$ and thus

$If c_{21}=1, c_{22}=1$

A general solution of the system is given as

$X(t)=Ac_{1}e^{\beta _{1}} + Bc_{2}e^{\beta _{2}t$ where the c's are the eigen vectors.

Thus the general solution is

$X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\$

Answer from: Quest

answer: segregation violated the equal protection clause of the 14th amendment, which holds that no state can “deny to any person within its jurisdiction the equal protection of the laws.”

details/explanation:

brown v. board of education of topeka, decided by the us supreme court in 1954, ruled that all americans are entitled to the same civil liberties and protections in regard to access to education. until that decision, it was legal to segregate schools according to race, so that black students could not attend the same schools as white students. an older supreme court decision, plessy v. ferguson (1896), had said that separate, segregated public facilities were acceptable as long as the facilities offered were equal in quality. in the case of brown v. board of education, that standard was challenged and defeated. segregation was shown to create inequality, and the supreme court unanimously ruled segregation to be unconstitutional.

the ruling was important in advancing civil rights because it affirmed that the 14th amendment applies to all rights and privileges of citizens, including access to education. this was being violated by states whose laws supported the segregation of schools. section 1 of the 14th amendment reads as follows:

all persons born or naturalized in the united states and subject to the jurisdiction thereof, are citizens of the united states and of the state wherein they reside. no state shall make or enforce any law which shall abridge the privileges or immunities of citizens of the united states; nor shall any state deprive any person of life, liberty, or property, without due process of law; nor deny to any person within its jurisdiction the equal protection of the laws.

Answer from: Quest

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Exercise 3.3.2: Verify that the system x® 0 1 3 3 1 x® has the two solutions 1 1 e 4t and 1 −1 e −2t...

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