Mixing
Independent solution means no repeated root.
$begingroup$
If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then
neither $y_1$ and nor $y_2$ has repeated zeros ?
They don't have common point of extremum.
I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with
ordinary-differential-equations
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
$endgroup$
add a comment |
$begingroup$
If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then
neither $y_1$ and nor $y_2$ has repeated zeros ?
They don't have common point of extremum.
I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with
ordinary-differential-equations
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
$endgroup$
add a comment |
$begingroup$
If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then
neither $y_1$ and nor $y_2$ has repeated zeros ?
They don't have common point of extremum.
I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with
ordinary-differential-equations
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
$endgroup$
If $y_1 $ and $y_2 $ are independent solution of homogeneous second order linear differential equation.Then
neither $y_1$ and nor $y_2$ has repeated zeros ?
They don't have common point of extremum.
I don't see why its true. I tried write it as linear combination in non trivial way but have no clue how to use hypothesis ( i will add more details if i found any) .please give me a hint to start with
ordinary-differential-equations
ordinary-differential-equations
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
asked Jan 29 at 18:55
Cloud JRCloud JR
910518
910518
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
for 2) how to use uniqueness, itsn't obvious for me
$endgroup$
– Cloud JR
Jan 30 at 13:55
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
for 2) how to use uniqueness, itsn't obvious for me
$endgroup$
– Cloud JR
Jan 30 at 13:55
add a comment |
$begingroup$
If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
$begingroup$
for 2) how to use uniqueness, itsn't obvious for me
$endgroup$
– Cloud JR
Jan 30 at 13:55
add a comment |
$begingroup$
If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
$endgroup$
If a "repeated zero" of $y_1$ is a point $x_0$ such that $y_1(x_0)=y_1'(x_0)=0$, then (1) is clear: Uniquenss for the IVP $ay''+by'+cy=0$, $y(x_0)=0$, $y'(x_0)=0$ shows that $y_1=0$, so $y_1$ and $y_2$ are not independent. Similarly for (2): If $y_1'(x_0)=y_2'(x_0)=0$ you can use uniqueness to show that $y_2$ is a multiple of $y_1$.
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
answered Jan 29 at 19:08
David C. UllrichDavid C. Ullrich
61.6k43995
61.6k43995
$begingroup$
for 2) how to use uniqueness, itsn't obvious for me
$endgroup$
– Cloud JR
Jan 30 at 13:55
add a comment |
$begingroup$
for 2) how to use uniqueness, itsn't obvious for me
$endgroup$
– Cloud JR
Jan 30 at 13:55
$begingroup$
for 2) how to use uniqueness, itsn't obvious for me
$endgroup$
– Cloud JR
Jan 30 at 13:55
$begingroup$
for 2) how to use uniqueness, itsn't obvious for me
$endgroup$
– Cloud JR
Jan 30 at 13:55
add a comment |
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