Question

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

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{t}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The Laplace transform of the given derivative equation $y^{\text{'}\text{'}}+9y=e^t,y\left(0\right)=0,y^{\text{'}}\left(0\right)=0$is $y\left(t\right)=\frac{1}{10}{e}^t-\frac{1}{10}\cos 3t-\frac{1}{30}\sin 3t$.

Step-by-step solution

Theorem of Transform of a Derivative.

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 $y^{\text{'}\text{'}}+9y=e^t$with the initial condition as

$\begin{array}{rcl}y\left(0\right)& =& 0,y^{\text{'}}\left(0\right)=0\\ L\left\{y^{\text{'}\text{'}}\right\}& =& s^{2}L\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)\to \left(1\right)\\ & & \end{array}$

Take the Laplace Transform on both the sides of the equation,

$\begin{array}{c}L\left\{e^t\right\}=L\left\{y^{\text{'}\text{'}}+9y\right\}\\ =L\left\{y^{\text{'}\text{'}}\right\}+9L\left\{y\right\}\end{array}$

Substitute the equation(1) here,

$\begin{array}{c}L\left\{e^t\right\}=s^{2}L\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)+9L\left\{y\right\}\\ =\left(s^2+9\right)L\left\{y\right\}\end{array}$

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

$\begin{array}{c}L\left\{e^t\right\}=\left(s^2+9\right)Y\left(s\right);\mathbf{since}\frac{1}{s-a}=L\left\{e^{at}\right\}\\ \frac{1}{s-1}=\left(s^2+9\right)Y\left(s\right)\\ Y\left(s\right)=\frac{1}{\left(s^2+9\right)\left(s-1\right)}\to \left(1\right)\end{array}$

Using partial fraction.

Decompose the fraction in the above equation using partial fraction technique,

$\begin{array}{l}\frac{1}{\left(s-1\right)\left(s^2+9\right)}=\frac{A}{s-1}+\frac{Bs+C}{s^2+9}\\ \frac{1}{\left(s-1\right)\left(s^2+9\right)}=\frac{A\left(s^2+9\right)+\left(Bs+C\right)\left(s-1\right)}{\left(s-1\right)\left(s^2+9\right)}\to \left(2\right)\end{array}$

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

$\begin{array}{l}1=A{s}^2+9A+B{s}^2-Bs+Cs-C\\ 1=\left(A+B\right)s^2+\left(C-B\right)s+9A-C\end{array}$

Comparing the coefficients on both sides, we get

$\begin{array}{l}A+B=0\\ C-B=0\\ 9A-C=1\\ A=\frac{1}{10},B=-\frac{1}{10},C=-\frac{1}{10}\end{array}$

Substitution of the partial fraction values.

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

$\begin{array}{c}\frac{1}{\left(s-1\right)\left(s^2+9\right)}=\frac{1}{10\left(s-1\right)}-\frac{s+1}{10\left(s^2+9\right)}\\ Y\left(s\right)=\frac{1}{10\left(s-1\right)}-\frac{s+1}{10\left(s^2+9\right)}\end{array}$

Finding the inverse Laplace Transform.

Now take the inverse Laplace Transform of both the sides of the equation,

$\begin{array}{c}{L}^{-1}\left\{Y\left(s\right)\right\}=L^{-1}\frac{1}{10\left(s-1\right)}-L^{-1}\frac{s+1}{10\left(s^2+9\right)}\\ L^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{10}{L}^{-1}\frac{1}{s-1}-\frac{1}{10}{L}^{-1}\frac{s}{s^2+3^2}-\frac{1}{10}{L}^{-1}\frac{\frac{3}{3}}{s^2+3^2}\\ L^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{10}{L}^{-1}\frac{1}{s-1}-\frac{1}{30}{L}^{-1}\frac{s}{s^2+3^2}-\frac{1}{30}{L}^{-1}\frac{3}{s^2+3^2}\\ since{L}^{-1}\frac{s}{s^2+k^2}=\cos kt;L^{-1}\frac{1}{s-a}=e^{at};L^{-1}\frac{k}{s^2+k^2}=\sin kt\\ L^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{10}{e}^t-\frac{1}{10}\cos 3t-\frac{1}{30}\sin 3t\\ y\left(t\right)=L^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{10}{e}^t-\frac{1}{10}\cos 3t-\frac{1}{30}\sin 3t\\ \end{array}$

Thus the Laplace Transform of the given derivative equation is $y\left(t\right)=\frac{1}{10}{e}^t-\frac{1}{10}\cos 3t-\frac{1}{30}\sin 3t$