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Laplace-Metodu-solution-y-5y-6y-cos-t-y-0-0-and-y-0-1-

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Question Number 150277 by mathdanisur last updated on 10/Aug/21
Laplace Metodu (solution) y^(′′) + 5y^′ + 6y = cos(t) y(0) = 0 and y^′ (0) = 1
$$\mathrm{Laplace}\:\mathrm{Metodu}\:\left(\mathrm{solution}\right) \\ $$$$\mathrm{y}^{''} \:+\:\mathrm{5y}^{'} \:+\:\mathrm{6y}\:=\:\mathrm{cos}\left(\mathrm{t}\right) \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$
Commented by amin96 last updated on 10/Aug/21
L{y′′}+5L{y′}+6L{y}=L{cos (t)} Y(s)s^2 −sy(0)−y′(0)+5(sY(s)−y(0))+6Y(s)=(s/(s^2 +1)) Y(s)s^2 −1+5sY(s)+6Y(s)=(s/(s^2 +1)) Y(s)(s^2 +5s+6)=(s/(s^2 +1))+1 Y(s)=(s/((s^2 +1)(s+2)(s+3)))+(1/((s+2)(s+3))) L^(−1) {Y(s)}=2L^(−1) {(1/(s+2))}−4L^(−1) {(1/(s^2 +1))}−L^(−1) {(1/(s+3))}+L^(−1) {(1/(s+2))}−L^(−1) {(1/(s+3))} y=2e^(−2t) −4sin t−e^(−3t) +e^(−2t) −e^(−3t) y=3e^(−2t) −2e^(−3t) −4sin t
$${L}\left\{{y}''\right\}+\mathrm{5}{L}\left\{{y}'\right\}+\mathrm{6}{L}\left\{{y}\right\}={L}\left\{\mathrm{cos}\:\left({t}\right)\right\} \\ $$$${Y}\left({s}\right){s}^{\mathrm{2}} −{sy}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)+\mathrm{5}\left({sY}\left({s}\right)−{y}\left(\mathrm{0}\right)\right)+\mathrm{6}{Y}\left({s}\right)=\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$$${Y}\left({s}\right){s}^{\mathrm{2}} −\mathrm{1}+\mathrm{5}{sY}\left({s}\right)+\mathrm{6}{Y}\left({s}\right)=\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$$${Y}\left({s}\right)\left({s}^{\mathrm{2}} +\mathrm{5}{s}+\mathrm{6}\right)=\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\: \\ $$$${Y}\left({s}\right)=\frac{{s}}{\left({s}^{\mathrm{2}} +\mathrm{1}\right)\left({s}+\mathrm{2}\right)\left({s}+\mathrm{3}\right)}+\frac{\mathrm{1}}{\left({s}+\mathrm{2}\right)\left({s}+\mathrm{3}\right)} \\ $$$${L}^{−\mathrm{1}} \left\{{Y}\left({s}\right)\right\}=\mathrm{2}{L}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{s}+\mathrm{2}}\right\}−\mathrm{4}{L}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{1}}\right\}−{L}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{s}+\mathrm{3}}\right\}+{L}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{s}+\mathrm{2}}\right\}−{L}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{s}+\mathrm{3}}\right\} \\ $$$${y}=\mathrm{2}{e}^{−\mathrm{2}{t}} −\mathrm{4sin}\:{t}−{e}^{−\mathrm{3}{t}} +{e}^{−\mathrm{2}{t}} −{e}^{−\mathrm{3}{t}} \\ $$$${y}=\mathrm{3}{e}^{−\mathrm{2}{t}} −\mathrm{2}{e}^{−\mathrm{3}{t}} −\mathrm{4sin}\:{t} \\ $$
Commented by mathdanisur last updated on 10/Aug/21
Ser, Thank You
$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{Thank}\:\mathrm{You} \\ $$

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