Laplace transform of $y'''+3y''+3y'=12e^{-t}$ where $y(0)=1, y'(0)=0, y''(0)=-3$












0












$begingroup$


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







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  • $begingroup$
    How about starting with partial fraction decomposition?
    $endgroup$
    – Michael Burr
    Dec 3 '18 at 1:54
















0












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    How about starting with partial fraction decomposition?
    $endgroup$
    – Michael Burr
    Dec 3 '18 at 1:54














0












0








0


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






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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


















  • $begingroup$
    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










1 Answer
1






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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.






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  • $begingroup$
    What do you think?
    $endgroup$
    – Emmanuel Misley
    Dec 3 '18 at 23:39











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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you think?
    $endgroup$
    – Emmanuel Misley
    Dec 3 '18 at 23:39
















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What do you think?
    $endgroup$
    – Emmanuel Misley
    Dec 3 '18 at 23:39














0












0








0





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








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


















  • $begingroup$
    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


















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