Laplace transform of $y'''+3y''+3y'=12e^{-t}$ where $y(0)=1, y'(0)=0, y''(0)=-3$
$begingroup$
I obtained
$frac{(s+4)(s^{2}+3)}{s(s+1)(s^{2}+3s+3)}$
Im not sure how to simplify $(s^{2}+3s+3)$ to use an inverse laplace.
I took the Laplace Transform of my equation and got the following.
$[s^{3}Y(s) - s^{2}y(0) - sy'(0) - y''(0)] +3[s^{2}Y(s) - sy(0) -y'(0)] + 3[sY(s)-y(0)]= frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]-s^{2}+3-3s-3=frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]=frac{(s+4)(s^2+3)}{s+1}$
Any help would be greatly appreciated.
ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
I obtained
$frac{(s+4)(s^{2}+3)}{s(s+1)(s^{2}+3s+3)}$
Im not sure how to simplify $(s^{2}+3s+3)$ to use an inverse laplace.
I took the Laplace Transform of my equation and got the following.
$[s^{3}Y(s) - s^{2}y(0) - sy'(0) - y''(0)] +3[s^{2}Y(s) - sy(0) -y'(0)] + 3[sY(s)-y(0)]= frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]-s^{2}+3-3s-3=frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]=frac{(s+4)(s^2+3)}{s+1}$
Any help would be greatly appreciated.
ordinary-differential-equations
$endgroup$
$begingroup$
How about starting with partial fraction decomposition?
$endgroup$
– Michael Burr
Dec 3 '18 at 1:54
add a comment |
$begingroup$
I obtained
$frac{(s+4)(s^{2}+3)}{s(s+1)(s^{2}+3s+3)}$
Im not sure how to simplify $(s^{2}+3s+3)$ to use an inverse laplace.
I took the Laplace Transform of my equation and got the following.
$[s^{3}Y(s) - s^{2}y(0) - sy'(0) - y''(0)] +3[s^{2}Y(s) - sy(0) -y'(0)] + 3[sY(s)-y(0)]= frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]-s^{2}+3-3s-3=frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]=frac{(s+4)(s^2+3)}{s+1}$
Any help would be greatly appreciated.
ordinary-differential-equations
$endgroup$
I obtained
$frac{(s+4)(s^{2}+3)}{s(s+1)(s^{2}+3s+3)}$
Im not sure how to simplify $(s^{2}+3s+3)$ to use an inverse laplace.
I took the Laplace Transform of my equation and got the following.
$[s^{3}Y(s) - s^{2}y(0) - sy'(0) - y''(0)] +3[s^{2}Y(s) - sy(0) -y'(0)] + 3[sY(s)-y(0)]= frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]-s^{2}+3-3s-3=frac{12}{(s+1)}$
$Y(s)[s^{3}+3s^{2}+3s]=frac{(s+4)(s^2+3)}{s+1}$
Any help would be greatly appreciated.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 3 '18 at 1:51
Andrea Zamora
asked Dec 3 '18 at 1:44
Andrea ZamoraAndrea Zamora
11
11
$begingroup$
How about starting with partial fraction decomposition?
$endgroup$
– Michael Burr
Dec 3 '18 at 1:54
add a comment |
$begingroup$
How about starting with partial fraction decomposition?
$endgroup$
– Michael Burr
Dec 3 '18 at 1:54
$begingroup$
How about starting with partial fraction decomposition?
$endgroup$
– Michael Burr
Dec 3 '18 at 1:54
$begingroup$
How about starting with partial fraction decomposition?
$endgroup$
– Michael Burr
Dec 3 '18 at 1:54
add a comment |
1 Answer
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$begingroup$
I don't know if it's correct but I tried this
$frac{(s+1)(s^2+3)}{s(s+1)(s^2+3s+3)}=frac{4}{s}-frac{12}{s+1}+frac{9s+15}{s^2+3s+3}$
and the first and second fractions
$mathcal{L}(4)=frac{4}{s} \ mathcal{L}(-12e^{-t})=-frac{12}{s+1}$
you can do this for the third one
$frac{9s+15}{s^2+3s+3} = frac{9s}{s^2+3s+3}+frac{15}{s^2+3s+3}=9 frac{s}{s^2+3s+3}+15 frac{1}{s^2+3s+3} $
let's take second fraction:
$frac{1}{s^{2}+3s+3}=frac{1}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{2}{sqrt{3}}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t))$
and for the first one
$frac{s+frac{3}{2}-frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{(s+frac{3}{2})}{(s+frac{3}{2})^{2}+frac{3}{4}}-frac{frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=mathcal{L}(e^{-frac{3}{2}t}cos(frac{sqrt{3}}{2}t))-sqrt{3}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t)) $
I'd do that. Sorry if I made some mistake.
$endgroup$
$begingroup$
What do you think?
$endgroup$
– Emmanuel Misley
Dec 3 '18 at 23:39
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I don't know if it's correct but I tried this
$frac{(s+1)(s^2+3)}{s(s+1)(s^2+3s+3)}=frac{4}{s}-frac{12}{s+1}+frac{9s+15}{s^2+3s+3}$
and the first and second fractions
$mathcal{L}(4)=frac{4}{s} \ mathcal{L}(-12e^{-t})=-frac{12}{s+1}$
you can do this for the third one
$frac{9s+15}{s^2+3s+3} = frac{9s}{s^2+3s+3}+frac{15}{s^2+3s+3}=9 frac{s}{s^2+3s+3}+15 frac{1}{s^2+3s+3} $
let's take second fraction:
$frac{1}{s^{2}+3s+3}=frac{1}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{2}{sqrt{3}}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t))$
and for the first one
$frac{s+frac{3}{2}-frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{(s+frac{3}{2})}{(s+frac{3}{2})^{2}+frac{3}{4}}-frac{frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=mathcal{L}(e^{-frac{3}{2}t}cos(frac{sqrt{3}}{2}t))-sqrt{3}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t)) $
I'd do that. Sorry if I made some mistake.
$endgroup$
$begingroup$
What do you think?
$endgroup$
– Emmanuel Misley
Dec 3 '18 at 23:39
add a comment |
$begingroup$
I don't know if it's correct but I tried this
$frac{(s+1)(s^2+3)}{s(s+1)(s^2+3s+3)}=frac{4}{s}-frac{12}{s+1}+frac{9s+15}{s^2+3s+3}$
and the first and second fractions
$mathcal{L}(4)=frac{4}{s} \ mathcal{L}(-12e^{-t})=-frac{12}{s+1}$
you can do this for the third one
$frac{9s+15}{s^2+3s+3} = frac{9s}{s^2+3s+3}+frac{15}{s^2+3s+3}=9 frac{s}{s^2+3s+3}+15 frac{1}{s^2+3s+3} $
let's take second fraction:
$frac{1}{s^{2}+3s+3}=frac{1}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{2}{sqrt{3}}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t))$
and for the first one
$frac{s+frac{3}{2}-frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{(s+frac{3}{2})}{(s+frac{3}{2})^{2}+frac{3}{4}}-frac{frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=mathcal{L}(e^{-frac{3}{2}t}cos(frac{sqrt{3}}{2}t))-sqrt{3}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t)) $
I'd do that. Sorry if I made some mistake.
$endgroup$
$begingroup$
What do you think?
$endgroup$
– Emmanuel Misley
Dec 3 '18 at 23:39
add a comment |
$begingroup$
I don't know if it's correct but I tried this
$frac{(s+1)(s^2+3)}{s(s+1)(s^2+3s+3)}=frac{4}{s}-frac{12}{s+1}+frac{9s+15}{s^2+3s+3}$
and the first and second fractions
$mathcal{L}(4)=frac{4}{s} \ mathcal{L}(-12e^{-t})=-frac{12}{s+1}$
you can do this for the third one
$frac{9s+15}{s^2+3s+3} = frac{9s}{s^2+3s+3}+frac{15}{s^2+3s+3}=9 frac{s}{s^2+3s+3}+15 frac{1}{s^2+3s+3} $
let's take second fraction:
$frac{1}{s^{2}+3s+3}=frac{1}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{2}{sqrt{3}}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t))$
and for the first one
$frac{s+frac{3}{2}-frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{(s+frac{3}{2})}{(s+frac{3}{2})^{2}+frac{3}{4}}-frac{frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=mathcal{L}(e^{-frac{3}{2}t}cos(frac{sqrt{3}}{2}t))-sqrt{3}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t)) $
I'd do that. Sorry if I made some mistake.
$endgroup$
I don't know if it's correct but I tried this
$frac{(s+1)(s^2+3)}{s(s+1)(s^2+3s+3)}=frac{4}{s}-frac{12}{s+1}+frac{9s+15}{s^2+3s+3}$
and the first and second fractions
$mathcal{L}(4)=frac{4}{s} \ mathcal{L}(-12e^{-t})=-frac{12}{s+1}$
you can do this for the third one
$frac{9s+15}{s^2+3s+3} = frac{9s}{s^2+3s+3}+frac{15}{s^2+3s+3}=9 frac{s}{s^2+3s+3}+15 frac{1}{s^2+3s+3} $
let's take second fraction:
$frac{1}{s^{2}+3s+3}=frac{1}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{2}{sqrt{3}}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t))$
and for the first one
$frac{s+frac{3}{2}-frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=frac{(s+frac{3}{2})}{(s+frac{3}{2})^{2}+frac{3}{4}}-frac{frac{3}{2}}{(s+frac{3}{2})^{2}+frac{3}{4}}=mathcal{L}(e^{-frac{3}{2}t}cos(frac{sqrt{3}}{2}t))-sqrt{3}mathcal{L}(e^{-frac{3}{2}t}sin(frac{sqrt{3}}{2}t)) $
I'd do that. Sorry if I made some mistake.
edited Dec 3 '18 at 4:34
answered Dec 3 '18 at 4:09
Emmanuel MisleyEmmanuel Misley
12
12
$begingroup$
What do you think?
$endgroup$
– Emmanuel Misley
Dec 3 '18 at 23:39
add a comment |
$begingroup$
What do you think?
$endgroup$
– Emmanuel Misley
Dec 3 '18 at 23:39
$begingroup$
What do you think?
$endgroup$
– Emmanuel Misley
Dec 3 '18 at 23:39
$begingroup$
What do you think?
$endgroup$
– Emmanuel Misley
Dec 3 '18 at 23:39
add a comment |
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$begingroup$
How about starting with partial fraction decomposition?
$endgroup$
– Michael Burr
Dec 3 '18 at 1:54