Question

In problems 21-30 use the Laplace transform to solve the given initial-value problem.

28. $\mathbf{2}\mathit{y}\mathbf{-}\mathbf{20}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{51}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given initial-value using the Laplace transform is

$y\left(t\right)=2e^{-5t}\cos \left(\sqrt{0.5}t\right)+\frac{10}{\sqrt{0.5}}{e}^{-5t}\sin \left(\sqrt{0.5}t\right)$

Step-by-step solution

Definition of Laplace transform

A specialtype of integral transform called the Laplace transform, posses the linearity property in solving linear initial-value problems.

Let f be a function defined for $t\ge 0$.

Then the integral$L\left\{f\left(t\right)=\right\}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt$ is said to be the Laplace transform of f, provided that the integral converges.

We have the form in,

$2y-20y\text{'}+51y=0.......y\left(0\right)=2,y\text{'}\left(0\right)=0$

Step 2: THEOREM 7.2.2 Transform of a Derivative

If$f,f\text{'},.....,f^{\left(n-1\right)}$are continuous on$[0,\infty )$and are of exponential order and if$f^{\left(n\right)}\left(t\right)$is piecewise continuous on$[0,\infty )$, then

$L\left\{f^{\left(n\right)}\left(t\right)\right\}=s^{n}F\left(s\right)-s^{n-1}f\left(0\right)-s^{n-2}f\text{'}\left(0\right)-...-f^{(n-1)}\left(0\right)$

Where $F\left(s\right)=L\left\{f\left(t\right)\right\}$.

Transforming derivatives:

role="math" localid="1664300104550" $L\left\{y\right\}=s^{2}Y-sy\left(0\right)-y\text{'}\left(0\right)$, and $L\left\{y\text{'}\right\}=sY-y\left(0\right)$

$L\left\{2y+20y\text{'}+51y\right\}=L\left\{0\right\}\to 2(s^{2}Y-s(2)-0)+20\left(sY-2\right)+51Y=0$

$\Rightarrow Y=\frac{4s+40}{2s^2+20s+51}$

After manipulating, we get

$\begin{array}{rcl}\frac{4s+40}{2s^2+20s+51}& =& \frac{4(s+10)}{2(s^2+10s+25.5)}\\ & & \begin{array}{c}=\frac{2(s+10)}{{\left(s+5\right)}^2+0.5}\\ =\frac{2s}{{\left(s+5\right)}^2+0.5}+\frac{20}{{\left(s+5\right)}^2+0.5}\end{array}\end{array}$

Step 3: THEOREM 7.3.1 First Translation Theorem

Simplify and factorizing, we get

$$\begin{gathered} \Rightarrow 2L^{-1}\left\{\frac{s+5-5}{{\left(s-(-5)\right)}^2+0.5}\right\}+20L^{-1}\left\{\frac{\sqrt{0.5}}{0.5}\frac{\sqrt{0.5}}{{\left(s-(-5)\right)}^2+0.5}\right\} \\ =2L^{-1}\left\{\frac{s+5}{{\left(s-(-5)\right)}^2+0.5}\right\}-2L^{-1}\left\{\frac{10.\sqrt{0.5}\sqrt{0.5}}{{\left(s-(-5)\right)}^2+0.5}\right\}+\frac{20}{\sqrt{0.5}}{L}^{-1}\left\{\frac{\sqrt{0.5}}{{\left(s-(-5)\right)}^2+0.5}\right\} \end{gathered}$$

$=2L^{-1}\left\{\underset{s\to s-(-5)}{\lim }\frac{s}{{\left(s\right)}^2+0.5}\right\}-20\sqrt{0.5}{L}^{-1}\left\{\underset{s\to s-(-5)}{\lim }\frac{\sqrt{0.5}}{{\left(s\right)}^2+0.5}\right\}+\frac{20}{\sqrt{0.5}}{L}^{-1}\left\{\underset{s\to s-(-5)}{\lim }\frac{\sqrt{0.5}}{{\left(s\right)}^2+0.5}\right\}$

THEOREM 7.3.1 First Translation Theorem

If $L\left\{f\left(t\right)\right\}=F\left(s\right)$and is any real number, then

$L\left\{e^{at}f\left(t\right)\right\}=F(s-a)$

$$\begin{gathered} y\left(t\right)=2e^{-5t}\cos \left(\sqrt{0.5}t\right)-20\sqrt{0.5}{e}^{-5t}\sin \left(\sqrt{0.5}t\right)+\frac{20}{\sqrt{0.5}}{e}^{-5t}\sin \left(\sqrt{0.5}t\right) \\ y\left(t\right)=2e^{-5t}\cos \left(\sqrt{0.5}t\right)+\frac{10}{\sqrt{0.5}}{e}^{-5t}\sin \left(\sqrt{0.5}t\right) \end{gathered}$$

Therefore, the solution for the given initial-value using the Laplace transform $2y-20y\text{'}+51y=0,y\left(0\right)=2,y\text{'}\left(0\right)=0$is $y\left(t\right)=2e^{-5t}\cos \left(\sqrt{0.5}t\right)+\frac{10}{\sqrt{0.5}}{e}^{-5t}\sin \left(\sqrt{0.5}t\right)$.