necessaryh

2021-02-14

$y"-8{y}^{\prime}+41y=0$

$y(0)=0$

${y}^{\prime}(0)=5$

Now, using Y for the Laplace transform of y(t), i.e., $Y=L\{y(t)\}$ find the equation that results from the differential equation's Laplace transformation.

Sadie Eaton

Skilled2021-02-15Added 104 answers

We have:

$y"-8{y}^{\prime}+41y=0$

$y(0)=0$

${y}^{\prime}(0)=5$

Solution:

Apply Laplace transform.

$L\left\{y"\right\}={s}^{2}Y(s)-sy(0)-{y}^{\prime}(0)$

$L\left\{{y}^{\prime}\right\}=sY(s)-y(0)$

$L\left\{y\right\}=Y(s)$

$L\left\{c\right\}=\frac{c}{s}$

$L\left\{y"\right\}-8L\left\{{y}^{\prime}\right\}+41L\left\{y\right\}=L\left\{0\right\}$

$\Rightarrow {s}^{2}Y(s)-sy(0)-{y}^{\prime}(0)-8[sY(s)-y(0)]+41Y(s)=\frac{0}{s}$

$\Rightarrow {s}^{2}Y(s)-0-5-8[sY(s)-0]+41Y(s)=0$

Thus, the equation is

${s}^{2}Y(s)-0-5-8[sY(s)-0]+41Y(s)=0$

Next,

${s}^{2}Y(s)-0-5-8[sY(s)-0]+41Y(s)=0$

$\Rightarrow Y(s)[{s}^{2}-8s+41]=5$

$\Rightarrow Y(s)=\frac{5}{({s}^{2}-8s+41)}$

Therefore, the value of $Y(s)=\frac{5}{({s}^{2}-8s+41)}$

Apply inverse Laplace transform

$Y(s)=\frac{5}{({s}^{2}-8s+41)}$

$=\frac{5}{({s}^{2}-8s+16+25)}$

$=\frac{5}{(s-4{)}^{2}+{5}^{2}}$

$y(t)={L}^{-1}\left(\frac{5}{(s-4{)}^{2}+{5}^{2}}\right)$

$\Rightarrow y(t)={e}^{4t}\mathrm{sin}(5t)$

Therefore, the value of $y(t)={e}^{4t}\mathrm{sin}(5t)$

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