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skelet666 [1.2K]
4 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]4 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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108-271+215-_____=-103
Im going to combine the first thee numbers and replace the blank with a variable, lets use x. 
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now I am going to subtract 52 on both sides to get the variable alone
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Im going to multiply both sides to eliminate the negitive in the variable
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If you want to check just plug 155 into the blank
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so after checking work the blank is equal to 155.


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