-b.1.3.12 .... is a solution of the DE

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The discussion confirms that the functions $y_2(t)=t^{-2}\ln t$ and $y_1(t)=t^{-2}$ are solutions to the differential equation $t^2y''+5ty'+4y=0$. The verification process involves calculating the first and second derivatives of $y_1(t)$, leading to a characteristic equation that simplifies to zero, thus validating the solution. Participants express that while the verification is straightforward, solving the equation can be complex and tedious.

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karush
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#12 hope I rewrote the problem ok

Verify that $y_2(t)=t^{-2}\ln t\quad y_1(t)=t^{-2}$ is a solution of $t^2y''+5ty'+4y=0$
think the first steop is to compose a charactistic equation using r
 

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$y_1(t) = t^{-2} \implies y'_1(t) = -2t^{-3} \implies y''_1(t) = 6t^{-4}$

$t^2 \cdot 6t^{-4} + 5t \cdot(-2t^{-3}) + 4 \cdot t^{-2} = 6t^{-2} - 10t^{-2} + 4t^{-2} = 0$

verified
 
so that is how it is done...
the examples were pretty mind numbing compared

I really think MHB should write textbooks...
 
karush said:
so that is how it is done...
the examples were pretty mind numbing compared

I really think MHB should write textbooks...
You are merely supposed to verify the solutions. So just plug them in. Solving the equation, on the other hand, can be a bit mind numbing until you get used to it.

-Dan
 

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