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skelet666 [1.2K]
3 years ago
5

Does the set {t, t Int} form a fundamental set of solutions for t^2y" -- ty' +y = 0?

Mathematics
1 answer:
Ivanshal [37]3 years ago
6 0

Answer:

yes

Step-by-step explanation:

We are given that a Cauchy Euler's equation

t^2y''-ty'+y=0 where t is not equal to zero

We are given that two solutions of given Cauchy Euler's equation are t,t ln t

We have to find  the solutions are independent or dependent.

To find  the solutions are independent or dependent we use wronskain

w(x)=\begin{vmatrix}y_1&y_2\\y'_1&y'_2\end{vmatrix}

If wrosnkian is not equal to zero then solutions are dependent and if wronskian is zero then the set of solution is independent.

Let y_1=t,y_2=t ln t

y'_1=1,y'_2=lnt+1

w(x)=\begin{vmatrix}t&t lnt\\1&lnt+1\end{vmatrix}

w(x)=t(lnt+1)-tlnt=tlnt+t-tlnt=t where t is not equal to zero.

Hence,the wronskian  is not equal to zero .Therefore, the set of solutions is independent.

Hence, the set {t , tln t} form a fundamental set of solutions for given equation.

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I could use some more help
kherson [118]

Answer:

\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]

Step-by-step explanation:

Given

A = \left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]

B = \left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]

Required

2A - 4B

To solve 2A - 4B, we first multiply matrix A by 2 and matrix B by 4

So, if

A = \left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]

2A = 2 *\left[\begin{array}{ccc}-1&9&2\\10&-10&2\\-5&6&-5\end{array}\right]

2A = \left[\begin{array}{ccc}-2&18&4\\20&-20&4\\-10&12&-10\end{array}\right]

If

B = \left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]

then

4B = 4*\left[\begin{array}{ccc}7&-1&-1\\6&-4&-3\\-10&10&-7\end{array}\right]

4B = \left[\begin{array}{ccc}28&-4&-4\\24&-16&-12\\-40&40&-28\end{array}\right]

So; 2A - 4B becomes

\left[\begin{array}{ccc}-2&18&4\\20&-20&4\\-10&12&-10\end{array}\right] - \left[\begin{array}{ccc}28&-4&-4\\24&-16&-12\\-40&40&-28\end{array}\right]

\left[\begin{array}{ccc}-2-28&18-(-4)&4-(-4)\\20-24&-20-(-16)&4-(-12)\\-10-(-40)&12-40&-10-(-28)\end{array}\right]

\left[\begin{array}{ccc}-30&18+4&4+4\\20-24&-20+16&4+12\\-10+40&12-40&-10+28\end{array}\right]

\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]

Hence, 2A - 4B is equivalent to

\left[\begin{array}{ccc}-30&22&8\\-4&-4&16\\30&-28&18\end{array}\right]

3 0
3 years ago
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Below are \triangle ABC△ABCtriangle, A, B, C and \triangle DEF△DEFtriangle, D, E, F. We assume that AB=DEAB=DEA, B, equals, D, E
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Kyle has a storage box that is 2 ft. long, 3 ft. high, and has a volume of 12 ft.3
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Alright, let get started.

Kyle has a storage box having dimensions 2 ft long, 3 ft high and volume 12 cubic feet.

The formula for volume is = length * width * height

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Suppose the width is w hence

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16 = 8 w

Dividing 8 in both sides

w = 2 feet

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