Question

Use the Laplace transform to solve the given initial-value problem

$\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{1}$

Short answer

The Laplace transform of the given derivative equation

$2y^{\text{'}\text{'}\text{'}}+3y^{\text{'}\text{'}}-3y^{\text{'}}-2y=e^{-t},y\left(0\right)=0,y^{\text{'}}\left(0\right)=0,y^{\text{'}\text{'}}\left(0\right)=1$

$y\left(t\right)=\frac{5}{18}{e}^t+\frac{1}{2}{e}^{-t}-\frac{8}{9}{e}^{-\frac{t}{2}}+\frac{1}{9}{e}^{-2t}$

Step-by-step solution

Theorem of Transform of a Derivative and applying the Laplace Transform in the given equation.

If$\mathit{f}\mathbf{,}{\mathit{f}}^{\mathbf{\text{'}}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{f}}^{\left(n-1\right)}$are continuous onrole="math" localid="1663971217118" $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ and are of exponential order and if ${\mathit{f}}^{\left(n\right)}\left(t\right)$is a piecewise continuous on data-custom-editor="chemistry" $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$, then

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

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

The given derivative equation is $2y^{\text{'}\text{'}\text{'}}+3y^{\text{'}\text{'}}-3y^{\text{'}}-2y=e^{-t}$with the initial condition as

$y\left(0\right)=0,y^{\text{'}}\left(0\right)=0,y^{\text{'}\text{'}}\left(0\right)=1$

Applying Laplace Transform

$\begin{array}{c}L{y}^{\text{'}\text{'}\text{'}}=s^{3}Y\left(s\right)-s^{2}y\left(0\right)-s{y}^1\left(0\right)-y^{\text{'}\text{'}}\left(0\right)\\ L{y}^{\text{'}\text{'}}=s^{2}Y\left(s\right)-sy\left(0\right)-y^{\text{'}}\left(0\right)\\ L{y}^{\text{'}}=sY\left(s\right)-y\left(0\right)\\ L\left\{y\right\}=Y\left(s\right)\\ L{e}^{-t}=\frac{1}{s+1}\end{array}$

Applying the given initial condition,

$\begin{array}{c}y\left(0\right)=0,y^{\text{'}}\left(0\right)=0,y^{\text{'}\text{'}}\left(0\right)=1\\ L{y}^{\text{'}\text{'}\text{'}}=s^{3}Y\left(s\right)-s^2\times 0-s\times 0-1\\ L{y}^{\text{'}\text{'}\text{'}}=s^{3}Y\left(s\right)-1\\ L{y}^{\text{'}\text{'}}=s^{2}Y\left(s\right)-s\times 0-0\\ L{y}^{\text{'}\text{'}}=s^{2}Y\left(s\right)\\ L{y}^{\text{'}}=sY\left(s\right)-0\\ L{y}^{\text{'}}=sY\left(s\right)\end{array}$

Substitution of the Laplace equation into the differential equation.

Substitute it into the differential equation,

$\begin{array}{c}2y^{\text{'}\text{'}\text{'}}+3y^{\text{'}\text{'}}-3y^{\text{'}}-2y=e^{-t}\\ 2\left(s^{3}Y\left(s\right)-1\right)+3s^{2}Y\left(s\right)-3sY\left(s\right)-2Y\left(s\right)=\frac{1}{s+1}\\ 2s^{3}Y\left(s\right)-2+3s^{2}Y\left(s\right)-3sY\left(s\right)-2Y\left(s\right)=\frac{1}{s+1}\\ Y\left(s\right)2s^3+3s^2-3s-2=\frac{1}{s+1}+2\\ Y\left(s\right)2s^3+3s^2-3s-2=\frac{2s+3}{s+1}\end{array}$

To factor $2s^3+3s^2-3s-2$ ,firstdetermine what factor of $2$make the expression zero.

Notice that $s=1$ gives us $(s-1)$as a factor. Now divide $(s-1)$into $2s^3+3s^2-3s-2$to find the remaining factor which $2s^2+5s+2or\left(2s+1\right)\left(s+2\right)$

$\begin{array}{c}Y\left(s\right)2s^3+3s^2-3s-2=\frac{2s+3}{s+1}\\ Y\left(s\right)\left[\left(s-1\right)\left(2s+1\right)\left(s+2\right)\right]=\frac{2s+3}{s+1}\\ Y\left(s\right)=\frac{2s+3}{\left(s+1\right)\left(s-1\right)\left(2s+1\right)\left(s+2\right)}\end{array}$

Application of the partial fraction decomposition method.

Apply partial fraction decomposition method here,

$\begin{array}{c}\frac{2s+3}{\left(s+1\right)\left(s-1\right)\left(2s+1\right)\left(s+2\right)}=\frac{A}{s-1}+\frac{B}{s+1}+\frac{C}{2s+1}+\frac{D}{s+2}\\ 2s+3=A\left(s+1\right)\left(2s+1\right)\left(s+2\right)+B\left(s-1\right)\left(2s+1\right)\left(s+2\right)+C\left(s-1\right)\left(s+1\right)\left(s+2\right)+D\left(s-1\right)\left(s+1\right)\left(2s+1\right)\end{array}$

When$s=1$, we get,

$\begin{array}{c}2\left(1\right)+3=A\left(1+1\right)\left(2+1\right)\left(1+2\right)\\ 5=18A\\ A=\frac{5}{18}\end{array}$

When$s=-1$, we get,

$\begin{array}{c}2\left(-1\right)+3=B\left(-1-1\right)\left(-2+1\right)\left(-1+2\right)\\ 1=2B\\ B=\frac{1}{2}\end{array}$

When $s=-\frac{1}{2}$, we get,

$\begin{array}{c}2\left(-\frac{1}{2}\right)+3=C\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}+1\right)\left(-\frac{1}{2}+2\right)\\ 2=C\left(\frac{-3}{2}\right)\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\\ 2=-\frac{9}{8}C\\ C=-\frac{16}{9}\end{array}$

When $s=-2$, we get,

$\begin{array}{c}2\left(-2\right)+3=D\left(-2-1\right)\left(-2+1\right)\left(-4+2\right)\\ -1=-9D\\ D=\frac{1}{9}\end{array}$

Taking the Inverse Laplace Transform.

Substitute the values of $A,B,C,D$in the above equation,

$\begin{array}{c}Y\left(s\right)=\frac{2s+3}{\left(s+1\right)\left(s-1\right)\left(2s+1\right)\left(s+2\right)}\\ Y\left(s\right)=\frac{5}{18}\times \frac{1}{s-1}+\frac{1}{2}\times \frac{1}{s+1}-\frac{16}{9}\times \frac{1}{2s+1}+\frac{1}{9}\times \frac{1}{s+2}\\ Y\left(s\right)=\frac{5}{18}\times \frac{1}{s-1}+\frac{1}{2}\times \frac{1}{s+1}-\frac{16}{9}\times \frac{1}{2\left(s+\frac{1}{2}\right)}+\frac{1}{9}\times \frac{1}{s+2}\\ Y\left(s\right)=\frac{5}{18}\times \frac{1}{s-1}+\frac{1}{2}\times \frac{1}{s+1}-\frac{8}{9}\times \frac{1}{\left(s+\frac{1}{2}\right)}+\frac{1}{9}\times \frac{1}{s+2}\end{array}$

Take Inverse Laplace Transform,

$\begin{array}{c}{L}^{-1}\left\{Y\left(s\right)\right\}=\frac{5}{18}{L}^{-1}\frac{1}{s-1}+\frac{1}{2}{L}^{-1}\frac{1}{s+1}-\frac{8}{9}{L}^{-1}\frac{1}{s+\frac{1}{2}}+\frac{1}{9}{L}^{-1}\frac{1}{s+2}\\ y\left(t\right)=\frac{5}{18}{e}^t+\frac{1}{2}{e}^{-t}-\frac{8}{9}{e}^{-\frac{1}{2}}+\frac{1}{9}{e}^{-2t}\end{array}$

Thus the Laplace Transform of the given derivative equation is

$y\left(t\right)=\frac{5}{18}{e}^t+\frac{1}{2}{e}^{-t}-\frac{8}{9}{e}^{-\frac{1}{2}}+\frac{1}{9}{e}^{-2t}$