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Using inverse Laplace transform to solve differential equation
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The differential equation is as follows-
$$fracd^2 xdt^2 + 5 fracdxdt + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = frac1(s-1)(s+3)(s+2)$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions-
$X(s) = frac1(s-1)(s+3)(s+2) = fracAs-1 + fracBs+3 + fracCs+2 $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $
I am stuck from here on how to carry on this partial fraction
Can I sub all s values to be 0 ?
For example
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $
differential-equations laplace-transform partial-fractions
asked Jul 22 at 20:59
user185692
1045
add a comment |Â
up vote
3
down vote
favorite
The differential equation is as follows-
$$fracd^2 xdt^2 + 5 fracdxdt + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = frac1(s-1)(s+3)(s+2)$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions-
$X(s) = frac1(s-1)(s+3)(s+2) = fracAs-1 + fracBs+3 + fracCs+2 $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $
I am stuck from here on how to carry on this partial fraction
Can I sub all s values to be 0 ?
For example
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $
differential-equations laplace-transform partial-fractions
asked Jul 22 at 20:59
user185692
1045
1
You can expand everything, and compare the coefficients of $s^2$,$s$,$1$ on both sides. Or you can even substitute some values for $s$.
– Botond
Jul 22 at 21:02
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The differential equation is as follows-
$$fracd^2 xdt^2 + 5 fracdxdt + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = frac1(s-1)(s+3)(s+2)$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions-
$X(s) = frac1(s-1)(s+3)(s+2) = fracAs-1 + fracBs+3 + fracCs+2 $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $
I am stuck from here on how to carry on this partial fraction
Can I sub all s values to be 0 ?
For example
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $
differential-equations laplace-transform partial-fractions
asked Jul 22 at 20:59
user185692
1045
The differential equation is as follows-
$$fracd^2 xdt^2 + 5 fracdxdt + 6x = e^t $$
I use laplace transform to make it to become - $$X(s) = frac1(s-1)(s+3)(s+2)$$
where $X(s)$ is the Laplace transform of $X(t)$
So now I am trying to find $X(t)$ using inverse transform.
From partial fractions-
$X(s) = frac1(s-1)(s+3)(s+2) = fracAs-1 + fracBs+3 + fracCs+2 $
Numerator - $ 1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3) $
I am stuck from here on how to carry on this partial fraction
Can I sub all s values to be 0 ?
For example
$1 = A(0+3)(0+2)$
$1= B(0-1)(0+2) $
$1 = C (0-1)(0+3) $
differential-equations laplace-transform partial-fractions
asked Jul 22 at 20:59
user185692
1045
asked Jul 22 at 20:59
user185692
1045
asked Jul 22 at 20:59
user185692
1045
asked Jul 22 at 20:59
user185692
1045
1045
1
You can expand everything, and compare the coefficients of $s^2$,$s$,$1$ on both sides. Or you can even substitute some values for $s$.
– Botond
Jul 22 at 21:02
add a comment |Â
1
You can expand everything, and compare the coefficients of $s^2$,$s$,$1$ on both sides. Or you can even substitute some values for $s$.
– Botond
Jul 22 at 21:02
1
1
You can expand everything, and compare the coefficients of $s^2$,$s$,$1$ on both sides. Or you can even substitute some values for $s$.
– Botond
Jul 22 at 21:02
You can expand everything, and compare the coefficients of $s^2$,$s$,$1$ on both sides. Or you can even substitute some values for $s$.
– Botond
Jul 22 at 21:02
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we substitute $s=1$ we get that
$$1=A(1+3)(1+2)$$
With $s=-2$ we get that
$$1=C(-2-1)(-2+3)$$
And with $s=-3$ we get that
$$1=B(-3-1)(-3+2)$$
And here is a video about the partial fraction decomposition, you might find it helpful.
answered Jul 22 at 21:15
Botond
3,8432632
how do you know what coefficients to put into A , B and C for the “A+B+C†, “5A+B+2C†, “6A-2B-3C†?
– user185692
Jul 22 at 22:02
@user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it?
– Botond
Jul 22 at 22:04
The way you have gotten the 3 equations.
– user185692
Jul 22 at 22:07
@user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now?
– Botond
Jul 22 at 22:11
Yes!! Thanks a lot ..
– user185692
Jul 22 at 22:32
 |Â
show 1 more comment
up vote
3
down vote
$ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 implies 12A=1 implies A=frac 112$$
$$ s=-3 implies 4B=1 implies B=frac 14$$
$$ s=-2 implies -3C=1 implies C=frac -13$$
Now you proceed with the inverse Laplace Transform.
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok?
– user185692
Jul 22 at 21:34
Or must I only identify -3 and use -3 ?
– user185692
Jul 22 at 21:35
@user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present.
– Mohammad Riazi-Kermani
Jul 22 at 21:39
Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!!
– user185692
Jul 22 at 21:48
@user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$.
– Mohammad Riazi-Kermani
Jul 22 at 21:51
add a comment |Â
up vote
1
down vote
$$X(s) = frac1(s-1)(s+3)(s+2)$$
Substitute $k=s+2$
$$F(k) = frac1(k-3)(k+1)k$$
$$F(k) = frac1(k-3)left(frac 1 (k+1)k right)$$
$$F(k) = frac1(k-3)left(frac 1 k-frac 1 k+1 right)$$
$$F(k) = frac1(k-3)frac 1 k-frac1(k-3)frac 1 (k+1) $$
$$F(k) = frac 13 left (frac1(k-3)-frac 1 kright )-frac 14left(frac1(k-3)-frac 1 (k+1) right )$$
$$F(k) = frac 1 12 frac1(k-3)-frac 13k+frac 14frac 1 (k+1) $$
Substitute back $k=s+2$
$$X(s)=-frac 13(s+2)+frac 14(s+3)+frac 1 12(s-1)$$
$$x(t)=-frac 13e^-2t+frac 14e^-3t+frac 1 12e^t$$
It's a bit longer but i find this method easier
answered Jul 22 at 21:31
Isham
10.6k3829
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we substitute $s=1$ we get that
$$1=A(1+3)(1+2)$$
With $s=-2$ we get that
$$1=C(-2-1)(-2+3)$$
And with $s=-3$ we get that
$$1=B(-3-1)(-3+2)$$
And here is a video about the partial fraction decomposition, you might find it helpful.
answered Jul 22 at 21:15
Botond
3,8432632
how do you know what coefficients to put into A , B and C for the “A+B+C†, “5A+B+2C†, “6A-2B-3C†?
– user185692
Jul 22 at 22:02
@user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it?
– Botond
Jul 22 at 22:04
The way you have gotten the 3 equations.
– user185692
Jul 22 at 22:07
@user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now?
– Botond
Jul 22 at 22:11
Yes!! Thanks a lot ..
– user185692
Jul 22 at 22:32
 |Â
show 1 more comment
up vote
2
down vote
accepted
So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we substitute $s=1$ we get that
$$1=A(1+3)(1+2)$$
With $s=-2$ we get that
$$1=C(-2-1)(-2+3)$$
And with $s=-3$ we get that
$$1=B(-3-1)(-3+2)$$
And here is a video about the partial fraction decomposition, you might find it helpful.
answered Jul 22 at 21:15
Botond
3,8432632
how do you know what coefficients to put into A , B and C for the “A+B+C†, “5A+B+2C†, “6A-2B-3C†?
– user185692
Jul 22 at 22:02
@user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it?
– Botond
Jul 22 at 22:04
The way you have gotten the 3 equations.
– user185692
Jul 22 at 22:07
@user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now?
– Botond
Jul 22 at 22:11
Yes!! Thanks a lot ..
– user185692
Jul 22 at 22:32
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we substitute $s=1$ we get that
$$1=A(1+3)(1+2)$$
With $s=-2$ we get that
$$1=C(-2-1)(-2+3)$$
And with $s=-3$ we get that
$$1=B(-3-1)(-3+2)$$
And here is a video about the partial fraction decomposition, you might find it helpful.
answered Jul 22 at 21:15
Botond
3,8432632
So, if we have that
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
We can expand everything to get
$$1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$$
$$1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$$
So we must have that $A+B+C=0$, $5A+B+2C=0$ and $6A-2B-3C=1$.
Or alternatively:
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
If we substitute $s=1$ we get that
$$1=A(1+3)(1+2)$$
With $s=-2$ we get that
$$1=C(-2-1)(-2+3)$$
And with $s=-3$ we get that
$$1=B(-3-1)(-3+2)$$
And here is a video about the partial fraction decomposition, you might find it helpful.
answered Jul 22 at 21:15
Botond
3,8432632
edited Jul 22 at 21:20
answered Jul 22 at 21:15
Botond
3,8432632
answered Jul 22 at 21:15
Botond
3,8432632
answered Jul 22 at 21:15
Botond
3,8432632
3,8432632
how do you know what coefficients to put into A , B and C for the “A+B+C†, “5A+B+2C†, “6A-2B-3C†?
– user185692
Jul 22 at 22:02
@user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it?
– Botond
Jul 22 at 22:04
The way you have gotten the 3 equations.
– user185692
Jul 22 at 22:07
@user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now?
– Botond
Jul 22 at 22:11
Yes!! Thanks a lot ..
– user185692
Jul 22 at 22:32
 |Â
show 1 more comment
how do you know what coefficients to put into A , B and C for the “A+B+C†, “5A+B+2C†, “6A-2B-3C†?
– user185692
Jul 22 at 22:02
@user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it?
– Botond
Jul 22 at 22:04
The way you have gotten the 3 equations.
– user185692
Jul 22 at 22:07
@user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now?
– Botond
Jul 22 at 22:11
Yes!! Thanks a lot ..
– user185692
Jul 22 at 22:32
how do you know what coefficients to put into A , B and C for the “A+B+C†, “5A+B+2C†, “6A-2B-3C†?
– user185692
Jul 22 at 22:02
how do you know what coefficients to put into A , B and C for the “A+B+C†, “5A+B+2C†, “6A-2B-3C†?
– user185692
Jul 22 at 22:02
@user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it?
– Botond
Jul 22 at 22:04
@user185692 What do you mean? Would you like to ask the way I've got the 3 equation or the way you can solve it?
– Botond
Jul 22 at 22:04
The way you have gotten the 3 equations.
– user185692
Jul 22 at 22:07
The way you have gotten the 3 equations.
– user185692
Jul 22 at 22:07
@user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now?
– Botond
Jul 22 at 22:11
@user185692 So, after expanding everything, we get that $1=A(s^2+5s+6)+B(s^2+s-2)+C(s^2+2s-3)$. Now we want to make the right-hand side nicer, to look like a "normal" polynomial, something like the $ax^2+bx+c$. And if I didn't mess it up, we get that $1=s^2(A+B+C)+s(5A+B+2C)+1(6A-2B-3C)$. The $2$ sides look really different, right? But they should be equal for every $s$! So how can it be? If the right-hand side looks exactly like the left-hand side. So, how many $s^2$ do we have on the right? $A+B+C$. And on the left? $0$. So they should be equal: $A+B+C=0$, and so on. Is it clear now?
– Botond
Jul 22 at 22:11
Yes!! Thanks a lot ..
– user185692
Jul 22 at 22:32
Yes!! Thanks a lot ..
– user185692
Jul 22 at 22:32
 |Â
show 1 more comment
up vote
3
down vote
$ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 implies 12A=1 implies A=frac 112$$
$$ s=-3 implies 4B=1 implies B=frac 14$$
$$ s=-2 implies -3C=1 implies C=frac -13$$
Now you proceed with the inverse Laplace Transform.
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok?
– user185692
Jul 22 at 21:34
Or must I only identify -3 and use -3 ?
– user185692
Jul 22 at 21:35
@user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present.
– Mohammad Riazi-Kermani
Jul 22 at 21:39
Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!!
– user185692
Jul 22 at 21:48
@user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$.
– Mohammad Riazi-Kermani
Jul 22 at 21:51
add a comment |Â
up vote
3
down vote
$ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 implies 12A=1 implies A=frac 112$$
$$ s=-3 implies 4B=1 implies B=frac 14$$
$$ s=-2 implies -3C=1 implies C=frac -13$$
Now you proceed with the inverse Laplace Transform.
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok?
– user185692
Jul 22 at 21:34
Or must I only identify -3 and use -3 ?
– user185692
Jul 22 at 21:35
@user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present.
– Mohammad Riazi-Kermani
Jul 22 at 21:39
Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!!
– user185692
Jul 22 at 21:48
@user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$.
– Mohammad Riazi-Kermani
Jul 22 at 21:51
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up vote
3
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up vote
3
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$ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 implies 12A=1 implies A=frac 112$$
$$ s=-3 implies 4B=1 implies B=frac 14$$
$$ s=-2 implies -3C=1 implies C=frac -13$$
Now you proceed with the inverse Laplace Transform.
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
$ s=0$ is not the best choice
There are three values to assign to $s$ which makes our life very easy.
$$1 = A(s+3)(s+2) + B(s-1)(s+2) + C(s-1)(s+3)$$
$$ s=1 implies 12A=1 implies A=frac 112$$
$$ s=-3 implies 4B=1 implies B=frac 14$$
$$ s=-2 implies -3C=1 implies C=frac -13$$
Now you proceed with the inverse Laplace Transform.
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
answered Jul 22 at 21:29
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok?
– user185692
Jul 22 at 21:34
Or must I only identify -3 and use -3 ?
– user185692
Jul 22 at 21:35
@user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present.
– Mohammad Riazi-Kermani
Jul 22 at 21:39
Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!!
– user185692
Jul 22 at 21:48
@user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$.
– Mohammad Riazi-Kermani
Jul 22 at 21:51
add a comment |Â
For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok?
– user185692
Jul 22 at 21:34
Or must I only identify -3 and use -3 ?
– user185692
Jul 22 at 21:35
@user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present.
– Mohammad Riazi-Kermani
Jul 22 at 21:39
Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!!
– user185692
Jul 22 at 21:48
@user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$.
– Mohammad Riazi-Kermani
Jul 22 at 21:51
For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok?
– user185692
Jul 22 at 21:34
For example , why can’t I use -4 to find B ? Apparently I will get B= 1/10 which is different if I use -3. is this ok?
– user185692
Jul 22 at 21:34
Or must I only identify -3 and use -3 ?
– user185692
Jul 22 at 21:35
Or must I only identify -3 and use -3 ?
– user185692
Jul 22 at 21:35
@user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present.
– Mohammad Riazi-Kermani
Jul 22 at 21:39
@user185692 If you use -4 then you will have $1=A(-4+3)(-4+2)+B(-4−1)(-4+2)+C(-4−1)(-4+3)$ which does not give you a simple solution for $B$. As you see $ A$ and $C$ are still present.
– Mohammad Riazi-Kermani
Jul 22 at 21:39
Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!!
– user185692
Jul 22 at 21:48
Why must I sub all the s values to be -4 whereas you sub 1,-3,-2 for A,B, and C respectively ? I was asking must these 3 numbers be fixed to be 1,-3,-2 ? Like for example, could I use 1,-4,-2 respectively for A, B, C. Thanks!!!
– user185692
Jul 22 at 21:48
@user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$.
– Mohammad Riazi-Kermani
Jul 22 at 21:51
@user185692 When you let $s=1$ the coefficients of $B$ and $C$ become $0$. The only survivor is $A$.
– Mohammad Riazi-Kermani
Jul 22 at 21:51
add a comment |Â
up vote
1
down vote
$$X(s) = frac1(s-1)(s+3)(s+2)$$
Substitute $k=s+2$
$$F(k) = frac1(k-3)(k+1)k$$
$$F(k) = frac1(k-3)left(frac 1 (k+1)k right)$$
$$F(k) = frac1(k-3)left(frac 1 k-frac 1 k+1 right)$$
$$F(k) = frac1(k-3)frac 1 k-frac1(k-3)frac 1 (k+1) $$
$$F(k) = frac 13 left (frac1(k-3)-frac 1 kright )-frac 14left(frac1(k-3)-frac 1 (k+1) right )$$
$$F(k) = frac 1 12 frac1(k-3)-frac 13k+frac 14frac 1 (k+1) $$
Substitute back $k=s+2$
$$X(s)=-frac 13(s+2)+frac 14(s+3)+frac 1 12(s-1)$$
$$x(t)=-frac 13e^-2t+frac 14e^-3t+frac 1 12e^t$$
It's a bit longer but i find this method easier
answered Jul 22 at 21:31
Isham
10.6k3829
add a comment |Â
up vote
1
down vote
$$X(s) = frac1(s-1)(s+3)(s+2)$$
Substitute $k=s+2$
$$F(k) = frac1(k-3)(k+1)k$$
$$F(k) = frac1(k-3)left(frac 1 (k+1)k right)$$
$$F(k) = frac1(k-3)left(frac 1 k-frac 1 k+1 right)$$
$$F(k) = frac1(k-3)frac 1 k-frac1(k-3)frac 1 (k+1) $$
$$F(k) = frac 13 left (frac1(k-3)-frac 1 kright )-frac 14left(frac1(k-3)-frac 1 (k+1) right )$$
$$F(k) = frac 1 12 frac1(k-3)-frac 13k+frac 14frac 1 (k+1) $$
Substitute back $k=s+2$
$$X(s)=-frac 13(s+2)+frac 14(s+3)+frac 1 12(s-1)$$
$$x(t)=-frac 13e^-2t+frac 14e^-3t+frac 1 12e^t$$
It's a bit longer but i find this method easier
answered Jul 22 at 21:31
Isham
10.6k3829
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$X(s) = frac1(s-1)(s+3)(s+2)$$
Substitute $k=s+2$
$$F(k) = frac1(k-3)(k+1)k$$
$$F(k) = frac1(k-3)left(frac 1 (k+1)k right)$$
$$F(k) = frac1(k-3)left(frac 1 k-frac 1 k+1 right)$$
$$F(k) = frac1(k-3)frac 1 k-frac1(k-3)frac 1 (k+1) $$
$$F(k) = frac 13 left (frac1(k-3)-frac 1 kright )-frac 14left(frac1(k-3)-frac 1 (k+1) right )$$
$$F(k) = frac 1 12 frac1(k-3)-frac 13k+frac 14frac 1 (k+1) $$
Substitute back $k=s+2$
$$X(s)=-frac 13(s+2)+frac 14(s+3)+frac 1 12(s-1)$$
$$x(t)=-frac 13e^-2t+frac 14e^-3t+frac 1 12e^t$$
It's a bit longer but i find this method easier
answered Jul 22 at 21:31
Isham
10.6k3829
$$X(s) = frac1(s-1)(s+3)(s+2)$$
Substitute $k=s+2$
$$F(k) = frac1(k-3)(k+1)k$$
$$F(k) = frac1(k-3)left(frac 1 (k+1)k right)$$
$$F(k) = frac1(k-3)left(frac 1 k-frac 1 k+1 right)$$
$$F(k) = frac1(k-3)frac 1 k-frac1(k-3)frac 1 (k+1) $$
$$F(k) = frac 13 left (frac1(k-3)-frac 1 kright )-frac 14left(frac1(k-3)-frac 1 (k+1) right )$$
$$F(k) = frac 1 12 frac1(k-3)-frac 13k+frac 14frac 1 (k+1) $$
Substitute back $k=s+2$
$$X(s)=-frac 13(s+2)+frac 14(s+3)+frac 1 12(s-1)$$
$$x(t)=-frac 13e^-2t+frac 14e^-3t+frac 1 12e^t$$
It's a bit longer but i find this method easier
answered Jul 22 at 21:31
Isham
10.6k3829
edited Jul 22 at 21:55
answered Jul 22 at 21:31
Isham
10.6k3829
answered Jul 22 at 21:31
Isham
10.6k3829
answered Jul 22 at 21:31
Isham
10.6k3829
10.6k3829
add a comment |Â
add a comment |Â
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1
You can expand everything, and compare the coefficients of $s^2$,$s$,$1$ on both sides. Or you can even substitute some values for $s$.
– Botond
Jul 22 at 21:02