Integrating Factors - definite with Indefinite integrals The Next CEO of Stack OverflowCauchy-Schwarz Inequality for Integrals for any two functions clarificationInteresing, hard limit with sum, involving $pi$Help with $int_t_0^t_0+hdelta(t),dt=hdeltaleft(t_0+frac12hright)$Lebesgue integration in one variableCan't find adequate change of variables $y=u^alphax^beta$ for this equation.How to evaluate: $int_0^1x^n-1(1-x)^n+1dx$ODEs explosion timesEvaluate $2iint_-1^1fracoperatornameLi_2(-ix^2)-operatornameLi_2(ix^2)x^2dx$, where $operatornameLi_2(s)$ is the polylogarithmChanging Limits of Integration Rule?Differentiating definite integral seems weird to me.
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Integrating Factors - definite with Indefinite integrals
The Next CEO of Stack OverflowCauchy-Schwarz Inequality for Integrals for any two functions clarificationInteresing, hard limit with sum, involving $pi$Help with $int_t_0^t_0+hdelta(t),dt=hdeltaleft(t_0+frac12hright)$Lebesgue integration in one variableCan't find adequate change of variables $y=u^alphax^beta$ for this equation.How to evaluate: $int_0^1x^n-1(1-x)^n+1dx$ODEs explosion timesEvaluate $2iint_-1^1fracoperatornameLi_2(-ix^2)-operatornameLi_2(ix^2)x^2dx$, where $operatornameLi_2(s)$ is the polylogarithmChanging Limits of Integration Rule?Differentiating definite integral seems weird to me.
$begingroup$
I am self-learning ODEs through https://www.simiode.org/resources/3732/download/Granero-Belinchon-CourseDEAndApplications.pdf?451f56d1cb99f142dee7c2d16251bfcb=1.
Consider $P'(t)=-alpha P(t)+f(t), P(0)=P_0in mathbbR$
Clearly, then $fracddt[e^int_0^talpha (s) dsP(t)] = e^int_0^talpha (s)dsf(t)$. However, their next step was to say that this is iff
$[e^int_0^talpha (s) dsP(t)] = int_0^te^int_0^ualpha (s)dsf(u)du+C$ as if the LHS was integrated as an indefinite integral and the RHS as a definite integral. Is this generally possible? If so, how?
integration ordinary-differential-equations
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
$endgroup$
|
show 1 more comment
$begingroup$
I am self-learning ODEs through https://www.simiode.org/resources/3732/download/Granero-Belinchon-CourseDEAndApplications.pdf?451f56d1cb99f142dee7c2d16251bfcb=1.
Consider $P'(t)=-alpha P(t)+f(t), P(0)=P_0in mathbbR$
Clearly, then $fracddt[e^int_0^talpha (s) dsP(t)] = e^int_0^talpha (s)dsf(t)$. However, their next step was to say that this is iff
$[e^int_0^talpha (s) dsP(t)] = int_0^te^int_0^ualpha (s)dsf(u)du+C$ as if the LHS was integrated as an indefinite integral and the RHS as a definite integral. Is this generally possible? If so, how?
integration ordinary-differential-equations
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
$endgroup$
$begingroup$
If you like you can add constants $C_1$ to the left and $C_2$ to the right, but you can bring $C_1$ over to make it $C_2-C_1$ on the right. That is the same as just calling it $C$ on the right.
$endgroup$
– Michael
Mar 28 at 2:47
$begingroup$
@Michael but is it possible to integrate one side under an indefinite integral whilst the other is definite?
$endgroup$
– BayesIsBae
Mar 28 at 2:49
$begingroup$
You seemed to have changed the question after my comment. As I recall you had a $+C$ somewhere...?
$endgroup$
– Michael
Mar 28 at 2:49
$begingroup$
@Michael My bad, there should be a $+C$ on the RHS as I just added
$endgroup$
– BayesIsBae
Mar 28 at 2:51
$begingroup$
In that case my first comment applies. It looks like $C=P(0)$.
$endgroup$
– Michael
Mar 28 at 2:52
|
show 1 more comment
$begingroup$
I am self-learning ODEs through https://www.simiode.org/resources/3732/download/Granero-Belinchon-CourseDEAndApplications.pdf?451f56d1cb99f142dee7c2d16251bfcb=1.
Consider $P'(t)=-alpha P(t)+f(t), P(0)=P_0in mathbbR$
Clearly, then $fracddt[e^int_0^talpha (s) dsP(t)] = e^int_0^talpha (s)dsf(t)$. However, their next step was to say that this is iff
$[e^int_0^talpha (s) dsP(t)] = int_0^te^int_0^ualpha (s)dsf(u)du+C$ as if the LHS was integrated as an indefinite integral and the RHS as a definite integral. Is this generally possible? If so, how?
integration ordinary-differential-equations
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
$endgroup$
I am self-learning ODEs through https://www.simiode.org/resources/3732/download/Granero-Belinchon-CourseDEAndApplications.pdf?451f56d1cb99f142dee7c2d16251bfcb=1.
Consider $P'(t)=-alpha P(t)+f(t), P(0)=P_0in mathbbR$
Clearly, then $fracddt[e^int_0^talpha (s) dsP(t)] = e^int_0^talpha (s)dsf(t)$. However, their next step was to say that this is iff
$[e^int_0^talpha (s) dsP(t)] = int_0^te^int_0^ualpha (s)dsf(u)du+C$ as if the LHS was integrated as an indefinite integral and the RHS as a definite integral. Is this generally possible? If so, how?
integration ordinary-differential-equations
integration ordinary-differential-equations
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
edited Mar 28 at 2:51
BayesIsBae
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
asked Mar 28 at 2:45
BayesIsBaeBayesIsBae
817
817
$begingroup$
If you like you can add constants $C_1$ to the left and $C_2$ to the right, but you can bring $C_1$ over to make it $C_2-C_1$ on the right. That is the same as just calling it $C$ on the right.
$endgroup$
– Michael
Mar 28 at 2:47
$begingroup$
@Michael but is it possible to integrate one side under an indefinite integral whilst the other is definite?
$endgroup$
– BayesIsBae
Mar 28 at 2:49
$begingroup$
You seemed to have changed the question after my comment. As I recall you had a $+C$ somewhere...?
$endgroup$
– Michael
Mar 28 at 2:49
$begingroup$
@Michael My bad, there should be a $+C$ on the RHS as I just added
$endgroup$
– BayesIsBae
Mar 28 at 2:51
$begingroup$
In that case my first comment applies. It looks like $C=P(0)$.
$endgroup$
– Michael
Mar 28 at 2:52
|
show 1 more comment
$begingroup$
If you like you can add constants $C_1$ to the left and $C_2$ to the right, but you can bring $C_1$ over to make it $C_2-C_1$ on the right. That is the same as just calling it $C$ on the right.
$endgroup$
– Michael
Mar 28 at 2:47
$begingroup$
@Michael but is it possible to integrate one side under an indefinite integral whilst the other is definite?
$endgroup$
– BayesIsBae
Mar 28 at 2:49
$begingroup$
You seemed to have changed the question after my comment. As I recall you had a $+C$ somewhere...?
$endgroup$
– Michael
Mar 28 at 2:49
$begingroup$
@Michael My bad, there should be a $+C$ on the RHS as I just added
$endgroup$
– BayesIsBae
Mar 28 at 2:51
$begingroup$
In that case my first comment applies. It looks like $C=P(0)$.
$endgroup$
– Michael
Mar 28 at 2:52
$begingroup$
If you like you can add constants $C_1$ to the left and $C_2$ to the right, but you can bring $C_1$ over to make it $C_2-C_1$ on the right. That is the same as just calling it $C$ on the right.
$endgroup$
– Michael
Mar 28 at 2:47
$begingroup$
If you like you can add constants $C_1$ to the left and $C_2$ to the right, but you can bring $C_1$ over to make it $C_2-C_1$ on the right. That is the same as just calling it $C$ on the right.
$endgroup$
– Michael
Mar 28 at 2:47
$begingroup$
@Michael but is it possible to integrate one side under an indefinite integral whilst the other is definite?
$endgroup$
– BayesIsBae
Mar 28 at 2:49
$begingroup$
@Michael but is it possible to integrate one side under an indefinite integral whilst the other is definite?
$endgroup$
– BayesIsBae
Mar 28 at 2:49
$begingroup$
You seemed to have changed the question after my comment. As I recall you had a $+C$ somewhere...?
$endgroup$
– Michael
Mar 28 at 2:49
$begingroup$
You seemed to have changed the question after my comment. As I recall you had a $+C$ somewhere...?
$endgroup$
– Michael
Mar 28 at 2:49
$begingroup$
@Michael My bad, there should be a $+C$ on the RHS as I just added
$endgroup$
– BayesIsBae
Mar 28 at 2:51
$begingroup$
@Michael My bad, there should be a $+C$ on the RHS as I just added
$endgroup$
– BayesIsBae
Mar 28 at 2:51
$begingroup$
In that case my first comment applies. It looks like $C=P(0)$.
$endgroup$
– Michael
Mar 28 at 2:52
$begingroup$
In that case my first comment applies. It looks like $C=P(0)$.
$endgroup$
– Michael
Mar 28 at 2:52
|
show 1 more comment
0
active
oldest
votes
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$begingroup$
If you like you can add constants $C_1$ to the left and $C_2$ to the right, but you can bring $C_1$ over to make it $C_2-C_1$ on the right. That is the same as just calling it $C$ on the right.
$endgroup$
– Michael
Mar 28 at 2:47
$begingroup$
@Michael but is it possible to integrate one side under an indefinite integral whilst the other is definite?
$endgroup$
– BayesIsBae
Mar 28 at 2:49
$begingroup$
You seemed to have changed the question after my comment. As I recall you had a $+C$ somewhere...?
$endgroup$
– Michael
Mar 28 at 2:49
$begingroup$
@Michael My bad, there should be a $+C$ on the RHS as I just added
$endgroup$
– BayesIsBae
Mar 28 at 2:51
$begingroup$
In that case my first comment applies. It looks like $C=P(0)$.
$endgroup$
– Michael
Mar 28 at 2:52