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Unsure of Inverse Laplace Transform for B/(A-s^2)
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up vote
1
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I'm currently solving a differential equation via Laplace transforms and am unsure on how to apply the reverse transform to one of the terms. I have $L[F(x)]=f(s)$ and currently trying to perform:
$$L^-1left[fracBA-s^2right]$$
I've tried to take the negative of this and try and match it with one in a table, but have been unsuccessful. I know the answer is
$$fracB e^-sqrtA(e^2sqrtA-A)2sqrtA$$
but cannot get the workings correctly (determined the answer from mathematica, but cant get the steps right).
Can anyone help with which transform to use in the table?
differential-equations laplace-transform mathematical-physics
edited Jul 27 at 21:03
Dylan
11.4k31026
asked Jul 27 at 16:12
MathHelper123
102
add a comment |Â
up vote
1
down vote
favorite
I'm currently solving a differential equation via Laplace transforms and am unsure on how to apply the reverse transform to one of the terms. I have $L[F(x)]=f(s)$ and currently trying to perform:
$$L^-1left[fracBA-s^2right]$$
I've tried to take the negative of this and try and match it with one in a table, but have been unsuccessful. I know the answer is
$$fracB e^-sqrtA(e^2sqrtA-A)2sqrtA$$
but cannot get the workings correctly (determined the answer from mathematica, but cant get the steps right).
Can anyone help with which transform to use in the table?
differential-equations laplace-transform mathematical-physics
edited Jul 27 at 21:03
Dylan
11.4k31026
asked Jul 27 at 16:12
MathHelper123
102
Do you have any conditions on $A?$
– Adrian Keister
Jul 27 at 16:15
No conditions, A is just strictly some arbitrary constant.
– MathHelper123
Jul 27 at 16:16
Note Isham found a sign error in my previous answer, which I've now corrected.
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm currently solving a differential equation via Laplace transforms and am unsure on how to apply the reverse transform to one of the terms. I have $L[F(x)]=f(s)$ and currently trying to perform:
$$L^-1left[fracBA-s^2right]$$
I've tried to take the negative of this and try and match it with one in a table, but have been unsuccessful. I know the answer is
$$fracB e^-sqrtA(e^2sqrtA-A)2sqrtA$$
but cannot get the workings correctly (determined the answer from mathematica, but cant get the steps right).
Can anyone help with which transform to use in the table?
differential-equations laplace-transform mathematical-physics
edited Jul 27 at 21:03
Dylan
11.4k31026
asked Jul 27 at 16:12
MathHelper123
102
I'm currently solving a differential equation via Laplace transforms and am unsure on how to apply the reverse transform to one of the terms. I have $L[F(x)]=f(s)$ and currently trying to perform:
$$L^-1left[fracBA-s^2right]$$
I've tried to take the negative of this and try and match it with one in a table, but have been unsuccessful. I know the answer is
$$fracB e^-sqrtA(e^2sqrtA-A)2sqrtA$$
but cannot get the workings correctly (determined the answer from mathematica, but cant get the steps right).
Can anyone help with which transform to use in the table?
differential-equations laplace-transform mathematical-physics
edited Jul 27 at 21:03
Dylan
11.4k31026
asked Jul 27 at 16:12
MathHelper123
102
edited Jul 27 at 21:03
Dylan
11.4k31026
edited Jul 27 at 21:03
Dylan
11.4k31026
edited Jul 27 at 21:03
Dylan
11.4k31026
11.4k31026
asked Jul 27 at 16:12
MathHelper123
102
asked Jul 27 at 16:12
MathHelper123
102
asked Jul 27 at 16:12
MathHelper123
102
102
Do you have any conditions on $A?$
– Adrian Keister
Jul 27 at 16:15
No conditions, A is just strictly some arbitrary constant.
– MathHelper123
Jul 27 at 16:16
Note Isham found a sign error in my previous answer, which I've now corrected.
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
Do you have any conditions on $A?$
– Adrian Keister
Jul 27 at 16:15
No conditions, A is just strictly some arbitrary constant.
– MathHelper123
Jul 27 at 16:16
Note Isham found a sign error in my previous answer, which I've now corrected.
– Adrian Keister
Jul 27 at 16:47
Do you have any conditions on $A?$
– Adrian Keister
Jul 27 at 16:15
Do you have any conditions on $A?$
– Adrian Keister
Jul 27 at 16:15
No conditions, A is just strictly some arbitrary constant.
– MathHelper123
Jul 27 at 16:16
No conditions, A is just strictly some arbitrary constant.
– MathHelper123
Jul 27 at 16:16
Note Isham found a sign error in my previous answer, which I've now corrected.
– Adrian Keister
Jul 27 at 16:47
Note Isham found a sign error in my previous answer, which I've now corrected.
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
We have three possibilities, depending on the sign of $A$. First, let's assume $A>0.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=B,mathcalL^-1left(frac1left(sqrtA-sright)left(sqrtA+sright)right) \
&=B,mathcalL^-1left(frac12sqrtAleft(frac1sqrtA+s+frac1sqrtA-sright)right) \
&=fracB2sqrtA,mathcalL^-1left(frac1sqrtA+s+frac1sqrtA-sright)\
&=fracB2sqrtAleft(e^-sqrtA,t-e^sqrtA,tright)\
&=-fracBsqrtA,sinhleft(sqrtA,tright).
endalign*
You probably have unit step functions in there, technically, though you might not need them.
Now let's suppose $A<0.$ Let $C^2=|A|.$ So $C=sqrtA.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=mathcalL^-1left(fracB-s^2right) \
&=-fracBC,mathcalL^-1left(fracCC^2+s^2right) \
&=-fracBC,sin(C,t) \
&=-fracBsqrtA,sinleft(sqrtA,tright).
endalign*
Finally, suppose $A=0.$ Then you have merely
beginalign*
mathcalL^-1left(fracBA-s^2right)&=-B,mathcalL^-1left(frac1s^2right) \
&=-B,t.
endalign*
So, your final answer can be written this way:
$$mathcalL^-1left(fracBA-s^2right)=begincases-dfracBsqrtA,sinhleft(sqrtA,tright),quad &A>0\ -B, t,quad &A=0 \-dfracBsqrt-A,sinleft(sqrt-A,tright),quad &A<0
endcases. $$
answered Jul 27 at 16:33
Adrian Keister
3,58321533
Thank you so much, this helps immensely!
– MathHelper123
Jul 27 at 16:34
Adrian you have a sign mistake ...it should be Sinh not cosh
– Isham
Jul 27 at 16:37
@Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent.
– Adrian Keister
Jul 27 at 16:39
@Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on..
– Adrian Keister
Jul 27 at 16:44
@Isham: Found it, thanks!
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
up vote
0
down vote
$$f(t)=mathcalL^-1(B/(A-s^2))$$
$$f(t)=frac B2sqrt AmathcalL^-1(frac 1(sqrt A-s)+frac 1(sqrt A+s))$$
$$f(t)=frac B2sqrt AmathcalL^-1(-frac 1(s-sqrt A)+frac 1(s+sqrt A))$$
Look at the table of lapalce transform
$$f(t)=frac B2sqrt A(-e^sqrt At+e^-sqrt At)$$
$$f(t)=-frac Bsqrt Asinh(sqrt At)$$
answered Jul 27 at 16:26
Isham
10.5k3829
For $A,Bne 0$ .....otherwise for $A=0 implies f(t)=-Bt$
– Isham
Jul 27 at 17:45
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We have three possibilities, depending on the sign of $A$. First, let's assume $A>0.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=B,mathcalL^-1left(frac1left(sqrtA-sright)left(sqrtA+sright)right) \
&=B,mathcalL^-1left(frac12sqrtAleft(frac1sqrtA+s+frac1sqrtA-sright)right) \
&=fracB2sqrtA,mathcalL^-1left(frac1sqrtA+s+frac1sqrtA-sright)\
&=fracB2sqrtAleft(e^-sqrtA,t-e^sqrtA,tright)\
&=-fracBsqrtA,sinhleft(sqrtA,tright).
endalign*
You probably have unit step functions in there, technically, though you might not need them.
Now let's suppose $A<0.$ Let $C^2=|A|.$ So $C=sqrtA.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=mathcalL^-1left(fracB-s^2right) \
&=-fracBC,mathcalL^-1left(fracCC^2+s^2right) \
&=-fracBC,sin(C,t) \
&=-fracBsqrtA,sinleft(sqrtA,tright).
endalign*
Finally, suppose $A=0.$ Then you have merely
beginalign*
mathcalL^-1left(fracBA-s^2right)&=-B,mathcalL^-1left(frac1s^2right) \
&=-B,t.
endalign*
So, your final answer can be written this way:
$$mathcalL^-1left(fracBA-s^2right)=begincases-dfracBsqrtA,sinhleft(sqrtA,tright),quad &A>0\ -B, t,quad &A=0 \-dfracBsqrt-A,sinleft(sqrt-A,tright),quad &A<0
endcases. $$
answered Jul 27 at 16:33
Adrian Keister
3,58321533
Thank you so much, this helps immensely!
– MathHelper123
Jul 27 at 16:34
Adrian you have a sign mistake ...it should be Sinh not cosh
– Isham
Jul 27 at 16:37
@Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent.
– Adrian Keister
Jul 27 at 16:39
@Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on..
– Adrian Keister
Jul 27 at 16:44
@Isham: Found it, thanks!
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
up vote
1
down vote
accepted
We have three possibilities, depending on the sign of $A$. First, let's assume $A>0.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=B,mathcalL^-1left(frac1left(sqrtA-sright)left(sqrtA+sright)right) \
&=B,mathcalL^-1left(frac12sqrtAleft(frac1sqrtA+s+frac1sqrtA-sright)right) \
&=fracB2sqrtA,mathcalL^-1left(frac1sqrtA+s+frac1sqrtA-sright)\
&=fracB2sqrtAleft(e^-sqrtA,t-e^sqrtA,tright)\
&=-fracBsqrtA,sinhleft(sqrtA,tright).
endalign*
You probably have unit step functions in there, technically, though you might not need them.
Now let's suppose $A<0.$ Let $C^2=|A|.$ So $C=sqrtA.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=mathcalL^-1left(fracB-s^2right) \
&=-fracBC,mathcalL^-1left(fracCC^2+s^2right) \
&=-fracBC,sin(C,t) \
&=-fracBsqrtA,sinleft(sqrtA,tright).
endalign*
Finally, suppose $A=0.$ Then you have merely
beginalign*
mathcalL^-1left(fracBA-s^2right)&=-B,mathcalL^-1left(frac1s^2right) \
&=-B,t.
endalign*
So, your final answer can be written this way:
$$mathcalL^-1left(fracBA-s^2right)=begincases-dfracBsqrtA,sinhleft(sqrtA,tright),quad &A>0\ -B, t,quad &A=0 \-dfracBsqrt-A,sinleft(sqrt-A,tright),quad &A<0
endcases. $$
answered Jul 27 at 16:33
Adrian Keister
3,58321533
Thank you so much, this helps immensely!
– MathHelper123
Jul 27 at 16:34
Adrian you have a sign mistake ...it should be Sinh not cosh
– Isham
Jul 27 at 16:37
@Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent.
– Adrian Keister
Jul 27 at 16:39
@Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on..
– Adrian Keister
Jul 27 at 16:44
@Isham: Found it, thanks!
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We have three possibilities, depending on the sign of $A$. First, let's assume $A>0.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=B,mathcalL^-1left(frac1left(sqrtA-sright)left(sqrtA+sright)right) \
&=B,mathcalL^-1left(frac12sqrtAleft(frac1sqrtA+s+frac1sqrtA-sright)right) \
&=fracB2sqrtA,mathcalL^-1left(frac1sqrtA+s+frac1sqrtA-sright)\
&=fracB2sqrtAleft(e^-sqrtA,t-e^sqrtA,tright)\
&=-fracBsqrtA,sinhleft(sqrtA,tright).
endalign*
You probably have unit step functions in there, technically, though you might not need them.
Now let's suppose $A<0.$ Let $C^2=|A|.$ So $C=sqrtA.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=mathcalL^-1left(fracB-s^2right) \
&=-fracBC,mathcalL^-1left(fracCC^2+s^2right) \
&=-fracBC,sin(C,t) \
&=-fracBsqrtA,sinleft(sqrtA,tright).
endalign*
Finally, suppose $A=0.$ Then you have merely
beginalign*
mathcalL^-1left(fracBA-s^2right)&=-B,mathcalL^-1left(frac1s^2right) \
&=-B,t.
endalign*
So, your final answer can be written this way:
$$mathcalL^-1left(fracBA-s^2right)=begincases-dfracBsqrtA,sinhleft(sqrtA,tright),quad &A>0\ -B, t,quad &A=0 \-dfracBsqrt-A,sinleft(sqrt-A,tright),quad &A<0
endcases. $$
answered Jul 27 at 16:33
Adrian Keister
3,58321533
We have three possibilities, depending on the sign of $A$. First, let's assume $A>0.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=B,mathcalL^-1left(frac1left(sqrtA-sright)left(sqrtA+sright)right) \
&=B,mathcalL^-1left(frac12sqrtAleft(frac1sqrtA+s+frac1sqrtA-sright)right) \
&=fracB2sqrtA,mathcalL^-1left(frac1sqrtA+s+frac1sqrtA-sright)\
&=fracB2sqrtAleft(e^-sqrtA,t-e^sqrtA,tright)\
&=-fracBsqrtA,sinhleft(sqrtA,tright).
endalign*
You probably have unit step functions in there, technically, though you might not need them.
Now let's suppose $A<0.$ Let $C^2=|A|.$ So $C=sqrtA.$ Then we have
beginalign*
mathcalL^-1left(fracBA-s^2right)&=mathcalL^-1left(fracB-s^2right) \
&=-fracBC,mathcalL^-1left(fracCC^2+s^2right) \
&=-fracBC,sin(C,t) \
&=-fracBsqrtA,sinleft(sqrtA,tright).
endalign*
Finally, suppose $A=0.$ Then you have merely
beginalign*
mathcalL^-1left(fracBA-s^2right)&=-B,mathcalL^-1left(frac1s^2right) \
&=-B,t.
endalign*
So, your final answer can be written this way:
$$mathcalL^-1left(fracBA-s^2right)=begincases-dfracBsqrtA,sinhleft(sqrtA,tright),quad &A>0\ -B, t,quad &A=0 \-dfracBsqrt-A,sinleft(sqrt-A,tright),quad &A<0
endcases. $$
answered Jul 27 at 16:33
Adrian Keister
3,58321533
edited Jul 27 at 16:46
answered Jul 27 at 16:33
Adrian Keister
3,58321533
answered Jul 27 at 16:33
Adrian Keister
3,58321533
answered Jul 27 at 16:33
Adrian Keister
3,58321533
3,58321533
Thank you so much, this helps immensely!
– MathHelper123
Jul 27 at 16:34
Adrian you have a sign mistake ...it should be Sinh not cosh
– Isham
Jul 27 at 16:37
@Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent.
– Adrian Keister
Jul 27 at 16:39
@Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on..
– Adrian Keister
Jul 27 at 16:44
@Isham: Found it, thanks!
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
Thank you so much, this helps immensely!
– MathHelper123
Jul 27 at 16:34
Adrian you have a sign mistake ...it should be Sinh not cosh
– Isham
Jul 27 at 16:37
@Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent.
– Adrian Keister
Jul 27 at 16:39
@Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on..
– Adrian Keister
Jul 27 at 16:44
@Isham: Found it, thanks!
– Adrian Keister
Jul 27 at 16:47
Thank you so much, this helps immensely!
– MathHelper123
Jul 27 at 16:34
Thank you so much, this helps immensely!
– MathHelper123
Jul 27 at 16:34
Adrian you have a sign mistake ...it should be Sinh not cosh
– Isham
Jul 27 at 16:37
Adrian you have a sign mistake ...it should be Sinh not cosh
– Isham
Jul 27 at 16:37
@Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent.
– Adrian Keister
Jul 27 at 16:39
@Isham: Mathematica disagrees with you, I think. I could well have made a sign error, but it's not apparent.
– Adrian Keister
Jul 27 at 16:39
@Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on..
– Adrian Keister
Jul 27 at 16:44
@Isham: Hmm. Looks like you're right. Mistake has to be in the partial fraction decomposition. Hang on..
– Adrian Keister
Jul 27 at 16:44
@Isham: Found it, thanks!
– Adrian Keister
Jul 27 at 16:47
@Isham: Found it, thanks!
– Adrian Keister
Jul 27 at 16:47
add a comment |Â
up vote
0
down vote
$$f(t)=mathcalL^-1(B/(A-s^2))$$
$$f(t)=frac B2sqrt AmathcalL^-1(frac 1(sqrt A-s)+frac 1(sqrt A+s))$$
$$f(t)=frac B2sqrt AmathcalL^-1(-frac 1(s-sqrt A)+frac 1(s+sqrt A))$$
Look at the table of lapalce transform
$$f(t)=frac B2sqrt A(-e^sqrt At+e^-sqrt At)$$
$$f(t)=-frac Bsqrt Asinh(sqrt At)$$
answered Jul 27 at 16:26
Isham
10.5k3829
For $A,Bne 0$ .....otherwise for $A=0 implies f(t)=-Bt$
– Isham
Jul 27 at 17:45
add a comment |Â
up vote
0
down vote
$$f(t)=mathcalL^-1(B/(A-s^2))$$
$$f(t)=frac B2sqrt AmathcalL^-1(frac 1(sqrt A-s)+frac 1(sqrt A+s))$$
$$f(t)=frac B2sqrt AmathcalL^-1(-frac 1(s-sqrt A)+frac 1(s+sqrt A))$$
Look at the table of lapalce transform
$$f(t)=frac B2sqrt A(-e^sqrt At+e^-sqrt At)$$
$$f(t)=-frac Bsqrt Asinh(sqrt At)$$
answered Jul 27 at 16:26
Isham
10.5k3829
For $A,Bne 0$ .....otherwise for $A=0 implies f(t)=-Bt$
– Isham
Jul 27 at 17:45
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$f(t)=mathcalL^-1(B/(A-s^2))$$
$$f(t)=frac B2sqrt AmathcalL^-1(frac 1(sqrt A-s)+frac 1(sqrt A+s))$$
$$f(t)=frac B2sqrt AmathcalL^-1(-frac 1(s-sqrt A)+frac 1(s+sqrt A))$$
Look at the table of lapalce transform
$$f(t)=frac B2sqrt A(-e^sqrt At+e^-sqrt At)$$
$$f(t)=-frac Bsqrt Asinh(sqrt At)$$
answered Jul 27 at 16:26
Isham
10.5k3829
$$f(t)=mathcalL^-1(B/(A-s^2))$$
$$f(t)=frac B2sqrt AmathcalL^-1(frac 1(sqrt A-s)+frac 1(sqrt A+s))$$
$$f(t)=frac B2sqrt AmathcalL^-1(-frac 1(s-sqrt A)+frac 1(s+sqrt A))$$
Look at the table of lapalce transform
$$f(t)=frac B2sqrt A(-e^sqrt At+e^-sqrt At)$$
$$f(t)=-frac Bsqrt Asinh(sqrt At)$$
answered Jul 27 at 16:26
Isham
10.5k3829
edited Jul 27 at 16:32
answered Jul 27 at 16:26
Isham
10.5k3829
answered Jul 27 at 16:26
Isham
10.5k3829
answered Jul 27 at 16:26
Isham
10.5k3829
10.5k3829
For $A,Bne 0$ .....otherwise for $A=0 implies f(t)=-Bt$
– Isham
Jul 27 at 17:45
add a comment |Â
For $A,Bne 0$ .....otherwise for $A=0 implies f(t)=-Bt$
– Isham
Jul 27 at 17:45
For $A,Bne 0$ .....otherwise for $A=0 implies f(t)=-Bt$
– Isham
Jul 27 at 17:45
For $A,Bne 0$ .....otherwise for $A=0 implies f(t)=-Bt$
– Isham
Jul 27 at 17:45
add a comment |Â
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Do you have any conditions on $A?$
– Adrian Keister
Jul 27 at 16:15
No conditions, A is just strictly some arbitrary constant.
– MathHelper123
Jul 27 at 16:16
Note Isham found a sign error in my previous answer, which I've now corrected.
– Adrian Keister
Jul 27 at 16:47