Mixing
Solving a differential equation with the Dirac-Delta function without Laplace transformations
$begingroup$
So I'm trying to solve the following differential equation:
$y''+3y'+2y=delta(t-1)$, $y(0)=0$, $y'(0)=0$. (where $delta$ is the Dirac's delta function)
Everything I've read in my textbook/online has solved these types of equations by taking the Laplace transformation, but our class hasn't covered Laplace transformations yet...anyone have any idea what I should do?
ordinary-differential-equations laplace-transform
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
asked Oct 7 '11 at 3:28
benben
3112
$endgroup$
add a comment |
$begingroup$
So I'm trying to solve the following differential equation:
$y''+3y'+2y=delta(t-1)$, $y(0)=0$, $y'(0)=0$. (where $delta$ is the Dirac's delta function)
Everything I've read in my textbook/online has solved these types of equations by taking the Laplace transformation, but our class hasn't covered Laplace transformations yet...anyone have any idea what I should do?
ordinary-differential-equations laplace-transform
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
asked Oct 7 '11 at 3:28
benben
3112
$endgroup$
1
$begingroup$
What did you go over in class - have you even seen these sorts of differential equations before? Were you assigned some reading material preceding the homework? Are these textbook questions that lie at the end of a chapter? It's perplexing that you'd be expected to solve an equation but not expected to learn about how to do it first...
$endgroup$
– anon
Oct 7 '11 at 3:36
$begingroup$
Well, in class we did this example: y''-y=delta(t-3). The homogeneous solutions were y=e^-t and y=e^t. He then deduced that the particular solution would be: [(e^t)/2]*integral from 0 to t ( (e^-s)*delta(s-3)ds ) - [e^(-t)/2]*integral from 0 to t( (e^s)*delta(s-3))ds From there, he concluded that y(t) would be e^(t-3)/2 - e^(3-t)/2 for all t>3. I guess what I'm confused on is how much of the partucluar solution came from the homogenous, and how much is just always going to be part of the particular solution
$endgroup$
– ben
Oct 7 '11 at 3:53
1
$begingroup$
And does the way he 'then deduced' the particular solution have anything to do with formulas you already have? How did he justify it to the class if not with either a derivation or a formula you're expected to know?
$endgroup$
– anon
Oct 7 '11 at 4:02
$begingroup$
Dear ben, you can use the formula given in this answer.
$endgroup$
– Pierre-Yves Gaillard
Oct 7 '11 at 15:36
add a comment |
$begingroup$
So I'm trying to solve the following differential equation:
$y''+3y'+2y=delta(t-1)$, $y(0)=0$, $y'(0)=0$. (where $delta$ is the Dirac's delta function)
Everything I've read in my textbook/online has solved these types of equations by taking the Laplace transformation, but our class hasn't covered Laplace transformations yet...anyone have any idea what I should do?
ordinary-differential-equations laplace-transform
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
asked Oct 7 '11 at 3:28
benben
3112
$endgroup$
So I'm trying to solve the following differential equation:
$y''+3y'+2y=delta(t-1)$, $y(0)=0$, $y'(0)=0$. (where $delta$ is the Dirac's delta function)
Everything I've read in my textbook/online has solved these types of equations by taking the Laplace transformation, but our class hasn't covered Laplace transformations yet...anyone have any idea what I should do?
ordinary-differential-equations laplace-transform
ordinary-differential-equations laplace-transform
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
asked Oct 7 '11 at 3:28
benben
3112
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
asked Oct 7 '11 at 3:28
benben
3112
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
edited Oct 7 '11 at 9:03
J. M. is not a mathematician
61.5k5152290
61.5k5152290
asked Oct 7 '11 at 3:28
benben
3112
asked Oct 7 '11 at 3:28
benben
3112
asked Oct 7 '11 at 3:28
benben
3112
3112
1
$begingroup$
What did you go over in class - have you even seen these sorts of differential equations before? Were you assigned some reading material preceding the homework? Are these textbook questions that lie at the end of a chapter? It's perplexing that you'd be expected to solve an equation but not expected to learn about how to do it first...
$endgroup$
– anon
Oct 7 '11 at 3:36
$begingroup$
Well, in class we did this example: y''-y=delta(t-3). The homogeneous solutions were y=e^-t and y=e^t. He then deduced that the particular solution would be: [(e^t)/2]*integral from 0 to t ( (e^-s)*delta(s-3)ds ) - [e^(-t)/2]*integral from 0 to t( (e^s)*delta(s-3))ds From there, he concluded that y(t) would be e^(t-3)/2 - e^(3-t)/2 for all t>3. I guess what I'm confused on is how much of the partucluar solution came from the homogenous, and how much is just always going to be part of the particular solution
$endgroup$
– ben
Oct 7 '11 at 3:53
1
$begingroup$
And does the way he 'then deduced' the particular solution have anything to do with formulas you already have? How did he justify it to the class if not with either a derivation or a formula you're expected to know?
$endgroup$
– anon
Oct 7 '11 at 4:02
$begingroup$
Dear ben, you can use the formula given in this answer.
$endgroup$
– Pierre-Yves Gaillard
Oct 7 '11 at 15:36
add a comment |
1
$begingroup$
What did you go over in class - have you even seen these sorts of differential equations before? Were you assigned some reading material preceding the homework? Are these textbook questions that lie at the end of a chapter? It's perplexing that you'd be expected to solve an equation but not expected to learn about how to do it first...
$endgroup$
– anon
Oct 7 '11 at 3:36
$begingroup$
Well, in class we did this example: y''-y=delta(t-3). The homogeneous solutions were y=e^-t and y=e^t. He then deduced that the particular solution would be: [(e^t)/2]*integral from 0 to t ( (e^-s)*delta(s-3)ds ) - [e^(-t)/2]*integral from 0 to t( (e^s)*delta(s-3))ds From there, he concluded that y(t) would be e^(t-3)/2 - e^(3-t)/2 for all t>3. I guess what I'm confused on is how much of the partucluar solution came from the homogenous, and how much is just always going to be part of the particular solution
$endgroup$
– ben
Oct 7 '11 at 3:53
1
$begingroup$
And does the way he 'then deduced' the particular solution have anything to do with formulas you already have? How did he justify it to the class if not with either a derivation or a formula you're expected to know?
$endgroup$
– anon
Oct 7 '11 at 4:02
$begingroup$
Dear ben, you can use the formula given in this answer.
$endgroup$
– Pierre-Yves Gaillard
Oct 7 '11 at 15:36
1
1
$begingroup$
What did you go over in class - have you even seen these sorts of differential equations before? Were you assigned some reading material preceding the homework? Are these textbook questions that lie at the end of a chapter? It's perplexing that you'd be expected to solve an equation but not expected to learn about how to do it first...
$endgroup$
– anon
Oct 7 '11 at 3:36
$begingroup$
What did you go over in class - have you even seen these sorts of differential equations before? Were you assigned some reading material preceding the homework? Are these textbook questions that lie at the end of a chapter? It's perplexing that you'd be expected to solve an equation but not expected to learn about how to do it first...
$endgroup$
– anon
Oct 7 '11 at 3:36
$begingroup$
Well, in class we did this example: y''-y=delta(t-3). The homogeneous solutions were y=e^-t and y=e^t. He then deduced that the particular solution would be: [(e^t)/2]*integral from 0 to t ( (e^-s)*delta(s-3)ds ) - [e^(-t)/2]*integral from 0 to t( (e^s)*delta(s-3))ds From there, he concluded that y(t) would be e^(t-3)/2 - e^(3-t)/2 for all t>3. I guess what I'm confused on is how much of the partucluar solution came from the homogenous, and how much is just always going to be part of the particular solution
$endgroup$
– ben
Oct 7 '11 at 3:53
$begingroup$
Well, in class we did this example: y''-y=delta(t-3). The homogeneous solutions were y=e^-t and y=e^t. He then deduced that the particular solution would be: [(e^t)/2]*integral from 0 to t ( (e^-s)*delta(s-3)ds ) - [e^(-t)/2]*integral from 0 to t( (e^s)*delta(s-3))ds From there, he concluded that y(t) would be e^(t-3)/2 - e^(3-t)/2 for all t>3. I guess what I'm confused on is how much of the partucluar solution came from the homogenous, and how much is just always going to be part of the particular solution
$endgroup$
– ben
Oct 7 '11 at 3:53
1
1
$begingroup$
And does the way he 'then deduced' the particular solution have anything to do with formulas you already have? How did he justify it to the class if not with either a derivation or a formula you're expected to know?
$endgroup$
– anon
Oct 7 '11 at 4:02
$begingroup$
And does the way he 'then deduced' the particular solution have anything to do with formulas you already have? How did he justify it to the class if not with either a derivation or a formula you're expected to know?
$endgroup$
– anon
Oct 7 '11 at 4:02
$begingroup$
Dear ben, you can use the formula given in this answer.
$endgroup$
– Pierre-Yves Gaillard
Oct 7 '11 at 15:36
$begingroup$
Dear ben, you can use the formula given in this answer.
$endgroup$
– Pierre-Yves Gaillard
Oct 7 '11 at 15:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The physicist's answer, not worrying about convergence, uniqueness, etc. It sounds like you have no problem away from $t=1$. Up to $1^-, y(t)=0$. Above $1$, you can solve it with $y$ being a sum of exponentials. Crossing $1$, you should integrate: $int_{1^-}^{1^+}y''+3y'=1=y'+3y|_{1^-}^{1^+}$ As $y$ can't change instantaneously, $y'$ has to go from $0$ to $1$. Then solve it starting at $t=1$ with $y=0, y'=1$
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Above 1, why is this case different from below 1? Aren't both the homogeneous case and are therefore, identical?
$endgroup$
– Christopher Turnbull
Nov 21 '16 at 3:52
$begingroup$
@ChristopherTurnbull: Yes, away from $1$ you are back to the homogeneous case, but the initial conditions are different. We had $y'(0)=0$, but $y'(1^+)=1$. That starts a damped oscillation.
$endgroup$
– Ross Millikan
Nov 21 '16 at 4:15
add a comment |
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1 Answer
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$begingroup$
The physicist's answer, not worrying about convergence, uniqueness, etc. It sounds like you have no problem away from $t=1$. Up to $1^-, y(t)=0$. Above $1$, you can solve it with $y$ being a sum of exponentials. Crossing $1$, you should integrate: $int_{1^-}^{1^+}y''+3y'=1=y'+3y|_{1^-}^{1^+}$ As $y$ can't change instantaneously, $y'$ has to go from $0$ to $1$. Then solve it starting at $t=1$ with $y=0, y'=1$
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Above 1, why is this case different from below 1? Aren't both the homogeneous case and are therefore, identical?
$endgroup$
– Christopher Turnbull
Nov 21 '16 at 3:52
$begingroup$
@ChristopherTurnbull: Yes, away from $1$ you are back to the homogeneous case, but the initial conditions are different. We had $y'(0)=0$, but $y'(1^+)=1$. That starts a damped oscillation.
$endgroup$
– Ross Millikan
Nov 21 '16 at 4:15
add a comment |
$begingroup$
The physicist's answer, not worrying about convergence, uniqueness, etc. It sounds like you have no problem away from $t=1$. Up to $1^-, y(t)=0$. Above $1$, you can solve it with $y$ being a sum of exponentials. Crossing $1$, you should integrate: $int_{1^-}^{1^+}y''+3y'=1=y'+3y|_{1^-}^{1^+}$ As $y$ can't change instantaneously, $y'$ has to go from $0$ to $1$. Then solve it starting at $t=1$ with $y=0, y'=1$
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Above 1, why is this case different from below 1? Aren't both the homogeneous case and are therefore, identical?
$endgroup$
– Christopher Turnbull
Nov 21 '16 at 3:52
$begingroup$
@ChristopherTurnbull: Yes, away from $1$ you are back to the homogeneous case, but the initial conditions are different. We had $y'(0)=0$, but $y'(1^+)=1$. That starts a damped oscillation.
$endgroup$
– Ross Millikan
Nov 21 '16 at 4:15
add a comment |
$begingroup$
The physicist's answer, not worrying about convergence, uniqueness, etc. It sounds like you have no problem away from $t=1$. Up to $1^-, y(t)=0$. Above $1$, you can solve it with $y$ being a sum of exponentials. Crossing $1$, you should integrate: $int_{1^-}^{1^+}y''+3y'=1=y'+3y|_{1^-}^{1^+}$ As $y$ can't change instantaneously, $y'$ has to go from $0$ to $1$. Then solve it starting at $t=1$ with $y=0, y'=1$
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
$endgroup$
The physicist's answer, not worrying about convergence, uniqueness, etc. It sounds like you have no problem away from $t=1$. Up to $1^-, y(t)=0$. Above $1$, you can solve it with $y$ being a sum of exponentials. Crossing $1$, you should integrate: $int_{1^-}^{1^+}y''+3y'=1=y'+3y|_{1^-}^{1^+}$ As $y$ can't change instantaneously, $y'$ has to go from $0$ to $1$. Then solve it starting at $t=1$ with $y=0, y'=1$
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
answered Oct 9 '11 at 3:28
Ross MillikanRoss Millikan
300k24200375
300k24200375
$begingroup$
Above 1, why is this case different from below 1? Aren't both the homogeneous case and are therefore, identical?
$endgroup$
– Christopher Turnbull
Nov 21 '16 at 3:52
$begingroup$
@ChristopherTurnbull: Yes, away from $1$ you are back to the homogeneous case, but the initial conditions are different. We had $y'(0)=0$, but $y'(1^+)=1$. That starts a damped oscillation.
$endgroup$
– Ross Millikan
Nov 21 '16 at 4:15
add a comment |
$begingroup$
Above 1, why is this case different from below 1? Aren't both the homogeneous case and are therefore, identical?
$endgroup$
– Christopher Turnbull
Nov 21 '16 at 3:52
$begingroup$
@ChristopherTurnbull: Yes, away from $1$ you are back to the homogeneous case, but the initial conditions are different. We had $y'(0)=0$, but $y'(1^+)=1$. That starts a damped oscillation.
$endgroup$
– Ross Millikan
Nov 21 '16 at 4:15
$begingroup$
Above 1, why is this case different from below 1? Aren't both the homogeneous case and are therefore, identical?
$endgroup$
– Christopher Turnbull
Nov 21 '16 at 3:52
$begingroup$
Above 1, why is this case different from below 1? Aren't both the homogeneous case and are therefore, identical?
$endgroup$
– Christopher Turnbull
Nov 21 '16 at 3:52
$begingroup$
@ChristopherTurnbull: Yes, away from $1$ you are back to the homogeneous case, but the initial conditions are different. We had $y'(0)=0$, but $y'(1^+)=1$. That starts a damped oscillation.
$endgroup$
– Ross Millikan
Nov 21 '16 at 4:15
$begingroup$
@ChristopherTurnbull: Yes, away from $1$ you are back to the homogeneous case, but the initial conditions are different. We had $y'(0)=0$, but $y'(1^+)=1$. That starts a damped oscillation.
$endgroup$
– Ross Millikan
Nov 21 '16 at 4:15
add a comment |
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1
$begingroup$
What did you go over in class - have you even seen these sorts of differential equations before? Were you assigned some reading material preceding the homework? Are these textbook questions that lie at the end of a chapter? It's perplexing that you'd be expected to solve an equation but not expected to learn about how to do it first...
$endgroup$
– anon
Oct 7 '11 at 3:36
$begingroup$
Well, in class we did this example: y''-y=delta(t-3). The homogeneous solutions were y=e^-t and y=e^t. He then deduced that the particular solution would be: [(e^t)/2]*integral from 0 to t ( (e^-s)*delta(s-3)ds ) - [e^(-t)/2]*integral from 0 to t( (e^s)*delta(s-3))ds From there, he concluded that y(t) would be e^(t-3)/2 - e^(3-t)/2 for all t>3. I guess what I'm confused on is how much of the partucluar solution came from the homogenous, and how much is just always going to be part of the particular solution
$endgroup$
– ben
Oct 7 '11 at 3:53
1
$begingroup$
And does the way he 'then deduced' the particular solution have anything to do with formulas you already have? How did he justify it to the class if not with either a derivation or a formula you're expected to know?
$endgroup$
– anon
Oct 7 '11 at 4:02
$begingroup$
Dear ben, you can use the formula given in this answer.
$endgroup$
– Pierre-Yves Gaillard
Oct 7 '11 at 15:36