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I found myself hard to understand how to proceed in an exercise given.

Suppose that $y_1 , y_2$ hold a fundamental set of solutions of : $y'' + p(t)y' + q(t)y = 0$ and also $y_3 = a_1y_1 + a_2y_2$ , $y_4 = b_1y_1 + b_2y_2$, where $a_1,a_2,b_1,b_2 \in \mathbb R$.

Show that : $W(y_3,y_4) = (a_1b_2 - a_2b_1)W(y_1,y_2)$

Do the solutions $y_3,y_4$ hold a fundamental set of solutions too ? Yes or not and why ?

Attempt/Think :

First of all I combined as $y = y_3 + y_4 = (a_1 + b_1)y_1 + (a_2 + b_2)y_2$ and said that it should be :

$(a_1 + b_1)y_1(t_0) + (a_2 + b_2)y_2(t_0) = y_0$ and $(a_1 + b_1)y_1'(t_0) + (a_2 + b_2)y_2'(t_0) = y_0'$

But after that, I got the same Wronski as always, since it's a combination of $y_1,y_2$, so I couldn't show the first part of the exercise.

As for the second, we know that a fundamental set of solutions is held if $W(y_3,y_4) \neq 0$. Now we know that $W(y_1,y_2) \neq 0$ but if $(a_1b_2 - a_2b_1) = 0$ then a fundamental set of solutions isn't held, so that's answered.

I would really appreciate any help in the first part of the exercise, because probably my mind has stuck in something simple. Thanks in advance !

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1 Answer 1

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Just insert into \begin{align} W(y_3,y_4)&=y_3y_4'-y_3'y_4=y_3(b_1y_1'+b_2y_2')-y_3'(b_1y_1+b_2y_2) \\&=b_1W(y_3,y_1)+b_2W(y_3,y_2) \\ W(y_3,y_1)&=… \\ W(y_3,y_2)&=… \end{align} etc.

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  • $\begingroup$ As I said, stuck in something simple. Thanks for the clear up ! $\endgroup$
    – Rebellos
    Nov 13, 2016 at 12:46

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