Help on the derivation of Abel's Theorem using the wronskian
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I'm finding it difficult to solve the first order differential equation in the proof to obtain the conclusion. I've solved 1st order ODE's before and not had problems and I feel really silly for not being able to do this. I know this is separable, but I can't get the solution in the required form.
$$
frac{{rm d}W}{{rm d}t} = -pW
$$
ordinary-differential-equations proof-explanation intuition
edited Dec 13 '18 at 1:43
caverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
I'm finding it difficult to solve the first order differential equation in the proof to obtain the conclusion. I've solved 1st order ODE's before and not had problems and I feel really silly for not being able to do this. I know this is separable, but I can't get the solution in the required form.
$$
frac{{rm d}W}{{rm d}t} = -pW
$$
ordinary-differential-equations proof-explanation intuition
edited Dec 13 '18 at 1:43
caverac
14.6k31130
$endgroup$
add a comment |
$begingroup$
I'm finding it difficult to solve the first order differential equation in the proof to obtain the conclusion. I've solved 1st order ODE's before and not had problems and I feel really silly for not being able to do this. I know this is separable, but I can't get the solution in the required form.
$$
frac{{rm d}W}{{rm d}t} = -pW
$$
ordinary-differential-equations proof-explanation intuition
edited Dec 13 '18 at 1:43
caverac
14.6k31130
$endgroup$
I'm finding it difficult to solve the first order differential equation in the proof to obtain the conclusion. I've solved 1st order ODE's before and not had problems and I feel really silly for not being able to do this. I know this is separable, but I can't get the solution in the required form.
$$
frac{{rm d}W}{{rm d}t} = -pW
$$
ordinary-differential-equations proof-explanation intuition
ordinary-differential-equations proof-explanation intuition
edited Dec 13 '18 at 1:43
caverac
14.6k31130
edited Dec 13 '18 at 1:43
caverac
14.6k31130
edited Dec 13 '18 at 1:43
caverac
14.6k31130
edited Dec 13 '18 at 1:43
caverac
14.6k31130
edited Dec 13 '18 at 1:43
caverac
14.6k31130
14.6k31130
asked Dec 13 '18 at 1:09
user503154
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Write the equation as
$$
frac{{rm d}W}{W} = -p {rm d}t
$$
Integrate both sides
$$
ln W = - int_{x_0}^x p(t){rm d}t + tilde{C}
$$
take the exponential
$$
W = Cexpleft(- int_{x_0}^x p(t){rm d}tright)
$$
where $C = e^{-tilde{C}} : {rm const}$
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you. Much appreciated.
$endgroup$
– user503154
Dec 13 '18 at 23:31
add a comment |
$begingroup$
This is probably an improvement of the answer by user caverac:
to avoid assuming that $W>0$ (which is required in that answer) multiply the equation $frac {dW} {dt} +pW=0$ by the integrating factor $e^{int_{x_0}^{x} p(t), dt}$ to get $frac d {dt} (We^{int_{x_0}^{x} p(t), dt})=0$. Hence $We^{int_{x_0}^{x} p(t), dt}$ is a constant $c$ and $W=ce^{-int_{x_0}^{x} p(t), dt}$. This avoids dividing by $W$ and taking logarithm of $W$.
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
$endgroup$
2
$begingroup$
Thank you very much
$endgroup$
– user503154
Dec 13 '18 at 23:30
1
$begingroup$
Certainly an improvement (+1)
$endgroup$
– caverac
Dec 14 '18 at 0:14
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write the equation as
$$
frac{{rm d}W}{W} = -p {rm d}t
$$
Integrate both sides
$$
ln W = - int_{x_0}^x p(t){rm d}t + tilde{C}
$$
take the exponential
$$
W = Cexpleft(- int_{x_0}^x p(t){rm d}tright)
$$
where $C = e^{-tilde{C}} : {rm const}$
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you. Much appreciated.
$endgroup$
– user503154
Dec 13 '18 at 23:31
add a comment |
$begingroup$
Write the equation as
$$
frac{{rm d}W}{W} = -p {rm d}t
$$
Integrate both sides
$$
ln W = - int_{x_0}^x p(t){rm d}t + tilde{C}
$$
take the exponential
$$
W = Cexpleft(- int_{x_0}^x p(t){rm d}tright)
$$
where $C = e^{-tilde{C}} : {rm const}$
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
$endgroup$
$begingroup$
Thank you. Much appreciated.
$endgroup$
– user503154
Dec 13 '18 at 23:31
add a comment |
$begingroup$
Write the equation as
$$
frac{{rm d}W}{W} = -p {rm d}t
$$
Integrate both sides
$$
ln W = - int_{x_0}^x p(t){rm d}t + tilde{C}
$$
take the exponential
$$
W = Cexpleft(- int_{x_0}^x p(t){rm d}tright)
$$
where $C = e^{-tilde{C}} : {rm const}$
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
$endgroup$
Write the equation as
$$
frac{{rm d}W}{W} = -p {rm d}t
$$
Integrate both sides
$$
ln W = - int_{x_0}^x p(t){rm d}t + tilde{C}
$$
take the exponential
$$
W = Cexpleft(- int_{x_0}^x p(t){rm d}tright)
$$
where $C = e^{-tilde{C}} : {rm const}$
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
answered Dec 13 '18 at 1:48
caveraccaverac
14.6k31130
14.6k31130
$begingroup$
Thank you. Much appreciated.
$endgroup$
– user503154
Dec 13 '18 at 23:31
add a comment |
$begingroup$
Thank you. Much appreciated.
$endgroup$
– user503154
Dec 13 '18 at 23:31
$begingroup$
Thank you. Much appreciated.
$endgroup$
– user503154
Dec 13 '18 at 23:31
$begingroup$
Thank you. Much appreciated.
$endgroup$
– user503154
Dec 13 '18 at 23:31
add a comment |
$begingroup$
This is probably an improvement of the answer by user caverac:
to avoid assuming that $W>0$ (which is required in that answer) multiply the equation $frac {dW} {dt} +pW=0$ by the integrating factor $e^{int_{x_0}^{x} p(t), dt}$ to get $frac d {dt} (We^{int_{x_0}^{x} p(t), dt})=0$. Hence $We^{int_{x_0}^{x} p(t), dt}$ is a constant $c$ and $W=ce^{-int_{x_0}^{x} p(t), dt}$. This avoids dividing by $W$ and taking logarithm of $W$.
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
$endgroup$
2
$begingroup$
Thank you very much
$endgroup$
– user503154
Dec 13 '18 at 23:30
1
$begingroup$
Certainly an improvement (+1)
$endgroup$
– caverac
Dec 14 '18 at 0:14
add a comment |
$begingroup$
This is probably an improvement of the answer by user caverac:
to avoid assuming that $W>0$ (which is required in that answer) multiply the equation $frac {dW} {dt} +pW=0$ by the integrating factor $e^{int_{x_0}^{x} p(t), dt}$ to get $frac d {dt} (We^{int_{x_0}^{x} p(t), dt})=0$. Hence $We^{int_{x_0}^{x} p(t), dt}$ is a constant $c$ and $W=ce^{-int_{x_0}^{x} p(t), dt}$. This avoids dividing by $W$ and taking logarithm of $W$.
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
$endgroup$
2
$begingroup$
Thank you very much
$endgroup$
– user503154
Dec 13 '18 at 23:30
1
$begingroup$
Certainly an improvement (+1)
$endgroup$
– caverac
Dec 14 '18 at 0:14
add a comment |
$begingroup$
This is probably an improvement of the answer by user caverac:
to avoid assuming that $W>0$ (which is required in that answer) multiply the equation $frac {dW} {dt} +pW=0$ by the integrating factor $e^{int_{x_0}^{x} p(t), dt}$ to get $frac d {dt} (We^{int_{x_0}^{x} p(t), dt})=0$. Hence $We^{int_{x_0}^{x} p(t), dt}$ is a constant $c$ and $W=ce^{-int_{x_0}^{x} p(t), dt}$. This avoids dividing by $W$ and taking logarithm of $W$.
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
$endgroup$
This is probably an improvement of the answer by user caverac:
to avoid assuming that $W>0$ (which is required in that answer) multiply the equation $frac {dW} {dt} +pW=0$ by the integrating factor $e^{int_{x_0}^{x} p(t), dt}$ to get $frac d {dt} (We^{int_{x_0}^{x} p(t), dt})=0$. Hence $We^{int_{x_0}^{x} p(t), dt}$ is a constant $c$ and $W=ce^{-int_{x_0}^{x} p(t), dt}$. This avoids dividing by $W$ and taking logarithm of $W$.
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
answered Dec 13 '18 at 5:55
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
58.5k42161
2
$begingroup$
Thank you very much
$endgroup$
– user503154
Dec 13 '18 at 23:30
1
$begingroup$
Certainly an improvement (+1)
$endgroup$
– caverac
Dec 14 '18 at 0:14
add a comment |
2
$begingroup$
Thank you very much
$endgroup$
– user503154
Dec 13 '18 at 23:30
1
$begingroup$
Certainly an improvement (+1)
$endgroup$
– caverac
Dec 14 '18 at 0:14
2
2
$begingroup$
Thank you very much
$endgroup$
– user503154
Dec 13 '18 at 23:30
$begingroup$
Thank you very much
$endgroup$
– user503154
Dec 13 '18 at 23:30
1
1
$begingroup$
Certainly an improvement (+1)
$endgroup$
– caverac
Dec 14 '18 at 0:14
$begingroup$
Certainly an improvement (+1)
$endgroup$
– caverac
Dec 14 '18 at 0:14
add a comment |
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