Confusion regarding the general solution theorem
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Theorem 1.13 (General Solutions)
If $y_1$ and $y_2$ are linearly independent solutions of the equation $L[y]=0$ on the interval $Isubsetmathbb R$ where $L[y]=y''+p(t)y'+q(t)y$ and $p(t), q(t)$ are continuous functions on $I$, then there exists unique constants $c_1, c_2$ such that every solution $y$ of the differential equation $L[y]=0$ can be written as a linear combination,
$$ y(t) = c_1y_1(t) + c_2y_2(t).$$
I'm confused on the part which says that there exists unique constants $c1,c2$ such that every solution $y$ of the DE can be written as a linear combination. Shouldn't it be that for every solution $y$ there exists unique constants?
I've only recall DE's that have a unique solution called the particular solution for unique $c1,c2$. I feel like I'm missing something.
ordinary-differential-equations intuition
edited Jan 26 at 22:39
Jean Marie
31.1k42155
$endgroup$
add a comment |
$begingroup$
Theorem 1.13 (General Solutions)
If $y_1$ and $y_2$ are linearly independent solutions of the equation $L[y]=0$ on the interval $Isubsetmathbb R$ where $L[y]=y''+p(t)y'+q(t)y$ and $p(t), q(t)$ are continuous functions on $I$, then there exists unique constants $c_1, c_2$ such that every solution $y$ of the differential equation $L[y]=0$ can be written as a linear combination,
$$ y(t) = c_1y_1(t) + c_2y_2(t).$$
I'm confused on the part which says that there exists unique constants $c1,c2$ such that every solution $y$ of the DE can be written as a linear combination. Shouldn't it be that for every solution $y$ there exists unique constants?
I've only recall DE's that have a unique solution called the particular solution for unique $c1,c2$. I feel like I'm missing something.
ordinary-differential-equations intuition
edited Jan 26 at 22:39
Jean Marie
31.1k42155
$endgroup$
4
$begingroup$
Yes, for every solution, there exist unique $c_1$ and $c_2$. Not the other way round
$endgroup$
– Shubham Johri
Dec 11 '18 at 20:26
add a comment |
$begingroup$
Theorem 1.13 (General Solutions)
If $y_1$ and $y_2$ are linearly independent solutions of the equation $L[y]=0$ on the interval $Isubsetmathbb R$ where $L[y]=y''+p(t)y'+q(t)y$ and $p(t), q(t)$ are continuous functions on $I$, then there exists unique constants $c_1, c_2$ such that every solution $y$ of the differential equation $L[y]=0$ can be written as a linear combination,
$$ y(t) = c_1y_1(t) + c_2y_2(t).$$
I'm confused on the part which says that there exists unique constants $c1,c2$ such that every solution $y$ of the DE can be written as a linear combination. Shouldn't it be that for every solution $y$ there exists unique constants?
I've only recall DE's that have a unique solution called the particular solution for unique $c1,c2$. I feel like I'm missing something.
ordinary-differential-equations intuition
edited Jan 26 at 22:39
Jean Marie
31.1k42155
$endgroup$
Theorem 1.13 (General Solutions)
If $y_1$ and $y_2$ are linearly independent solutions of the equation $L[y]=0$ on the interval $Isubsetmathbb R$ where $L[y]=y''+p(t)y'+q(t)y$ and $p(t), q(t)$ are continuous functions on $I$, then there exists unique constants $c_1, c_2$ such that every solution $y$ of the differential equation $L[y]=0$ can be written as a linear combination,
$$ y(t) = c_1y_1(t) + c_2y_2(t).$$
I'm confused on the part which says that there exists unique constants $c1,c2$ such that every solution $y$ of the DE can be written as a linear combination. Shouldn't it be that for every solution $y$ there exists unique constants?
I've only recall DE's that have a unique solution called the particular solution for unique $c1,c2$. I feel like I'm missing something.
ordinary-differential-equations intuition
ordinary-differential-equations intuition
edited Jan 26 at 22:39
Jean Marie
31.1k42155
edited Jan 26 at 22:39
Jean Marie
31.1k42155
edited Jan 26 at 22:39
Jean Marie
31.1k42155
edited Jan 26 at 22:39
Jean Marie
31.1k42155
edited Jan 26 at 22:39
Jean Marie
31.1k42155
31.1k42155
asked Dec 11 '18 at 20:21
user503154
4
$begingroup$
Yes, for every solution, there exist unique $c_1$ and $c_2$. Not the other way round
$endgroup$
– Shubham Johri
Dec 11 '18 at 20:26
add a comment |
4
$begingroup$
Yes, for every solution, there exist unique $c_1$ and $c_2$. Not the other way round
$endgroup$
– Shubham Johri
Dec 11 '18 at 20:26
4
4
$begingroup$
Yes, for every solution, there exist unique $c_1$ and $c_2$. Not the other way round
$endgroup$
– Shubham Johri
Dec 11 '18 at 20:26
$begingroup$
Yes, for every solution, there exist unique $c_1$ and $c_2$. Not the other way round
$endgroup$
– Shubham Johri
Dec 11 '18 at 20:26
add a comment |
1 Answer
1
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$begingroup$
You are not missing anything. The statement should have been
" Every solution is a linear combination of $y_1$ and $y_2$ which means for each solution y(t),there are constants $c_1$ and $c_2$ such that $y(t)= c_1 y_1(t) + c_2 y_2(t)$"
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
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1 Answer
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1 Answer
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$begingroup$
You are not missing anything. The statement should have been
" Every solution is a linear combination of $y_1$ and $y_2$ which means for each solution y(t),there are constants $c_1$ and $c_2$ such that $y(t)= c_1 y_1(t) + c_2 y_2(t)$"
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
You are not missing anything. The statement should have been
" Every solution is a linear combination of $y_1$ and $y_2$ which means for each solution y(t),there are constants $c_1$ and $c_2$ such that $y(t)= c_1 y_1(t) + c_2 y_2(t)$"
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
add a comment |
$begingroup$
You are not missing anything. The statement should have been
" Every solution is a linear combination of $y_1$ and $y_2$ which means for each solution y(t),there are constants $c_1$ and $c_2$ such that $y(t)= c_1 y_1(t) + c_2 y_2(t)$"
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
You are not missing anything. The statement should have been
" Every solution is a linear combination of $y_1$ and $y_2$ which means for each solution y(t),there are constants $c_1$ and $c_2$ such that $y(t)= c_1 y_1(t) + c_2 y_2(t)$"
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 11 '18 at 20:58
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
add a comment |
add a comment |
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Yes, for every solution, there exist unique $c_1$ and $c_2$. Not the other way round
$endgroup$
– Shubham Johri
Dec 11 '18 at 20:26