Inverse Laplace of $frac{s}{s^2+4s+5}$
0
$begingroup$
Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$
I obtained...
$$frac{s}{s^2+4s+4-4+5}$$
$$frac{s}{s^2+4s+1}$$
$$frac{s}{(s+2)^2+1}$$
I am not sure what my next step is?
ordinary-differential-equations
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
$endgroup$
add a comment |
0
$begingroup$
Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$
I obtained...
$$frac{s}{s^2+4s+4-4+5}$$
$$frac{s}{s^2+4s+1}$$
$$frac{s}{(s+2)^2+1}$$
I am not sure what my next step is?
ordinary-differential-equations
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
$endgroup$
add a comment |
0
0
0
$begingroup$
Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$
I obtained...
$$frac{s}{s^2+4s+4-4+5}$$
$$frac{s}{s^2+4s+1}$$
$$frac{s}{(s+2)^2+1}$$
I am not sure what my next step is?
ordinary-differential-equations
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
$endgroup$
Find the inverse Laplace Transformation of $frac{s}{s^2+4s+5}$
I obtained...
$$frac{s}{s^2+4s+4-4+5}$$
$$frac{s}{s^2+4s+1}$$
$$frac{s}{(s+2)^2+1}$$
I am not sure what my next step is?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
edited Dec 7 '18 at 5:50
Eevee Trainer
5,7961936
5,7961936
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
asked Dec 7 '18 at 5:48
sumthatup11sumthatup11
11
11
add a comment |
add a comment |
1 Answer
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oldest
votes
5
$begingroup$
$$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$
Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$
$$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$begingroup$
See mathworld.wolfram.com/LaplaceTransform.html
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 5:52
$begingroup$
How did you get s = s+2 -2? Why did you add 2?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:00
$begingroup$
@sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 6:02
$begingroup$
I do not understand your step in the numerator. I do not understand how s = s+2....?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:06
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
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oldest
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votes
5
$begingroup$
$$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$
Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$
$$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$begingroup$
See mathworld.wolfram.com/LaplaceTransform.html
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 5:52
$begingroup$
How did you get s = s+2 -2? Why did you add 2?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:00
$begingroup$
@sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 6:02
$begingroup$
I do not understand your step in the numerator. I do not understand how s = s+2....?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:06
add a comment |
5
$begingroup$
$$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$
Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$
$$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$begingroup$
See mathworld.wolfram.com/LaplaceTransform.html
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 5:52
$begingroup$
How did you get s = s+2 -2? Why did you add 2?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:00
$begingroup$
@sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 6:02
$begingroup$
I do not understand your step in the numerator. I do not understand how s = s+2....?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:06
add a comment |
5
5
5
$begingroup$
$$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$
Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$
$$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
$endgroup$
$$dfrac s{(s+2)^2+1^2}=dfrac{s+2}{(s+2)^2+1^2}-2cdotdfrac1{(s+2)^2+1^2}$$
Now use $$Lleft(e^{at}cos btright)=dfrac{s-a}{(s-a)^2+b^2}$$
$$Lleft(e^{at}sin btright)=dfrac b{(s-a)^2+b^2}$$
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
answered Dec 7 '18 at 5:51
lab bhattacharjeelab bhattacharjee
225k15157275
225k15157275
$begingroup$
See mathworld.wolfram.com/LaplaceTransform.html
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 5:52
$begingroup$
How did you get s = s+2 -2? Why did you add 2?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:00
$begingroup$
@sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 6:02
$begingroup$
I do not understand your step in the numerator. I do not understand how s = s+2....?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:06
add a comment |
$begingroup$
See mathworld.wolfram.com/LaplaceTransform.html
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 5:52
$begingroup$
How did you get s = s+2 -2? Why did you add 2?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:00
$begingroup$
@sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 6:02
$begingroup$
I do not understand your step in the numerator. I do not understand how s = s+2....?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:06
$begingroup$
See mathworld.wolfram.com/LaplaceTransform.html
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 5:52
$begingroup$
See mathworld.wolfram.com/LaplaceTransform.html
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 5:52
$begingroup$
How did you get s = s+2 -2? Why did you add 2?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:00
$begingroup$
How did you get s = s+2 -2? Why did you add 2?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:00
$begingroup$
@sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 6:02
$begingroup$
@sumthatup11, To utilize the two formulas, write $$s=A(s+2)+Bcdot(1)=s(A)+(A+B)implies A=1, 2A+B=0iff B=?$$
$endgroup$
– lab bhattacharjee
Dec 7 '18 at 6:02
$begingroup$
I do not understand your step in the numerator. I do not understand how s = s+2....?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:06
$begingroup$
I do not understand your step in the numerator. I do not understand how s = s+2....?
$endgroup$
– sumthatup11
Dec 7 '18 at 6:06
add a comment |
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