Question

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

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

Short answer

The Laplace transform of the given derivative equation $y^{\text{'}\text{'}\text{'}}+2y^{\text{'}\text{'}}-y^{\text{'}}-2y=\sin 3t,y\left(0\right)=0,y^{\text{'}}\left(0\right)=0,y^{\text{'}\text{'}}\left(0\right)=1$is

$y\left(t\right)=\frac{3}{130}\cos 3t-\frac{1}{65}\sin 3t+\frac{16}{39}{e}^{-2t}+\frac{13}{60}{e}^t-\frac{13}{20}{e}^{-t}$

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 on$\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$and are of exponential order and if${\mathit{f}}^{\left(n\right)}\left(t\right)$is a piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$, 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$y^{\text{'}\text{'}\text{'}}+2y^{\text{'}\text{'}}-y^{\text{'}}-2y=\sin 3t$with the initial condition as

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

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

Applying Laplace Transform

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

Denote $L\left\{y\right\}=Y\left(s\right)$ and replace in the above equation

$\begin{array}{c}L\left\{\sin 3t\right\}=\left(s^3+2s^2-s-2\right)Y\left(s\right)-1;\sin ce\frac{k}{s^2+k^2}=L\left\{\sin kt\right\}\\ \frac{3}{s^2+3^2}=\left(s^3+2s^2-s-2\right)Y\left(s\right)-1\\ \left(s^3+2s^2-s-2\right)Y\left(s\right)=\frac{3}{s^2+3^2}+1\\ \left(s^2\left(s+2\right)-\left(s+2\right)\right)Y\left(s\right)=\frac{3+s^2+9}{s^2+9}\\ \left(s+2\right)\left(s^2-1\right)Y\left(s\right)=\frac{3+s^2+9}{s^2+9}\\ \left(s+2\right)\left(s-1\right)\left(s+1\right)Y\left(s\right)=\frac{3+s^2+9}{s^2+9}\\ Y\left(s\right)=\frac{3+s^2+9}{\left(s^2+9\right)\left(s+2\right)\left(s-1\right)\left(s+1\right)}\to \left(1\right)\end{array}$

Application of the partial fraction decomposition method.

Decompose the fraction $\frac{3+s^2+9}{\left(s^2+9\right)\left(s+2\right)\left(s-1\right)\left(s+1\right)}$using the partial fraction technique,

$\begin{array}{l}\frac{3+s^2+9}{\left(s^2+9\right)\left(s+2\right)\left(s-1\right)\left(s+1\right)}=\frac{As+B}{s^2+9}+\frac{C}{s+2}+\frac{D}{s-1}+\frac{F}{s+1}\to \left(2\right)\\ \frac{3+s^2+9}{\left(s^2+9\right)\left(s+2\right)\left(s-1\right)\left(s+1\right)}=\frac{\left(As+B\right)\left(s+2\right)\left(s^2-1^2\right)+C\left(s^2+9\right)\left(s^2-1^2\right)+D\left(s^2+9\right)\left(s+2\right)\left(s+1\right)+F\left(s^2+9\right)\left(s+2\right)\left(s-1\right)}{\left(s^2+9\right)\left(s+2\right)\left(s-1\right)\left(s+1\right)}\\ \end{array}$

Since the fractions in the above equation has the same denominator, it takes their numerator to be equal,

$\begin{array}{l}3+s^2+9=\left(As+B\right)\left(s+2\right)\left(s^2-1^2\right)+C\left(s^2+9\right)\left(s^2-1^2\right)\\ +D\left(s^2+9\right)\left(s+2\right)\left(s+1\right)+F\left(s^2+9\right)\left(s+2\right)\left(s-1\right)\to \left(3\right)\end{array}$

When $s=1$, we get,

$\begin{array}{c}{1}^2+12=0+0+D\left(1+9\right)\left(1+2\right)\left(1+1\right)+0\\ 13=60D\\ D=\frac{13}{60}\end{array}$

When $s=-1$, we get,

$\begin{array}{c}{\left(-1\right)}^2+12=0+0+0+F{\left(-1\right)}^2+9\left(-1-1\right)\left(-1+2\right)\\ 13=-20F\\ F=-\frac{13}{20}\end{array}$

When role="math" localid="1663972861581" $s=-2$, we get,

$\begin{array}{c}{\left(-2\right)}^2+12=0+C{\left(-2\right)}^2+9\left(-2-1\right)\left(-2+1\right)+0+0\\ 16=39C\\ C=\frac{16}{39}\end{array}$

When $s=0$, we get,

$\begin{array}{c}12=-2B-9C+18D-18F\\ 12=-2B-9\frac{16}{39}+18\frac{13}{60}-18\frac{-13}{60}\\ B=-\frac{3}{65}\end{array}$

Substitute the values of $B,C,D,F$in equation (3), to obtain$A=\frac{3}{130}$

Taking the Inverse Laplace Transform.

Substitute the values of$A,B,C,D$ in the equation (2),

$\begin{array}{l}\frac{3+s^2+9}{\left(s^2+9\right)\left(s+2\right)\left(s-1\right)\left(s+1\right)}=\frac{\frac{3}{130}s-\frac{3}{65}}{s^2+9}+\frac{16}{\left(39\right)s+2}+\frac{13}{\left(60\right)s-1}-\frac{13}{\left(20\right)s+1}\\ \end{array}$

The equation (1) becomes,

$\begin{array}{l}Y\left(s\right)=\frac{\frac{3}{130}s-\frac{3}{65}}{s^2+9}+\frac{16}{39\left(s+2\right)}+\frac{13}{60\left(s-1\right)}-\frac{13}{20\left(s+1\right)}\\ \end{array}$

Take Inverse Laplace Transform,

$\begin{array}{rcl}{L}^{-1}\left\{Y\left(s\right)\right\}& =& \frac{3}{130}{L}^{-1}\frac{s}{s^2+3^2}-\frac{1}{65}{L}^{-1}\frac{3}{s^2+3^2}-\frac{16}{39}{L}^{-1}\frac{1}{s-\left(-2\right)}+\frac{13}{60}{L}^{-1}\frac{1}{s-1}-\frac{13}{20}{L}^{-1}\frac{1}{s-\left(-1\right)}\\ \sin ce{L}^{-1}\frac{s}{s^2+k^2}& =& \cos kt;L^{-1}\frac{1}{s-a}=e^{at};L^{-1}\frac{k}{s^2+k^2}=\sin kt\\ y\left(t\right)& =& \frac{3}{130}\cos 3t-\frac{1}{65}\sin 3t+\frac{16}{39}{e}^{-2t}+\frac{13}{60}{e}^t-\frac{13}{20}{e}^{-t}\\ & & \end{array}$

Thus the Laplace Transform of the given derivative equation is

$y\left(t\right)=\frac{3}{130}\cos 3t-\frac{1}{65}\sin 3t+\frac{16}{39}{e}^{-2t}+\frac{13}{60}{e}^t-\frac{13}{20}{e}^{-t}$