Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathit{t}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

Short answer

The solution of above initial-value problem is,

$y\left(t\right)=\frac{2}{27}+\frac{1}{9}t-\frac{2}{27}{e}^{3t}+\frac{10}{9}t{e}^{3t}$

Step-by-step solution

Step 1:

Note that we know how to find Laplace transform; it is time to use to solve differential equations. The Laplace transform of the derivative of a function is an algebraic expression rather than a differential expression.

Lety be continuous withy’ piecewise continuous. Also suppose that,

$y<K{e}^{at}$

Forsomepositiveconstantkandconstanta,

Then

$\begin{array}{l}L\left\{y\text{'}\right\}=sY\left(s\right)-y\left(0\right)\\ L\left\{y\text{'}\text{'}\right\}=s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)\end{array}$

Second translation theorem:

If $F\left(s\right)=L\left\{f\left(t\right)\right\}$and, $a>0$then

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

Where role="math" localid="1664238908985" $\mu \left(t-a\right)$is ,

$L\left\{\mu \left(t-a\right)\right\}=\frac{e^{-as}}{s}$

Step 2:

TakingLaplace transform of each member of differential equation:

$\begin{array}{l}y\text{'}\text{'}-6y\text{'}+9y=t,y\left(0\right)=0,y\text{'}\left(0\right)=1\\ L\left\{y\text{'}\text{'}\right\}-6L\left\{y\text{'}\right\}+9L\left\{y\right\}=L\left\{t\right\}\end{array}$

We get;

$s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)-6sY\left(s\right)+6y\left(0\right)+9Y\left(s\right)=\frac{1}{s^2}$

Rearranging the term and writing the equation in terms of and using the initial value to get.

$\begin{array}{l}\left(s^2-6s+9\right)Y\left(s\right)=\frac{1}{s^2}\\ Y\left(s\right)=\frac{1}{{\left(s-3\right)}^2}+\frac{1}{s^2{\left(s-3\right)}^2}\end{array}$

Step 3:

Takingthe inverse Laplace transform and using the property of linearity to get

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

Equating numerators, we get

$\begin{array}{l}1=s{\left(s-3\right)}^{2}A+B{\left(s-3\right)}^2+s^2\left(s-3\right)C+D{s}^2\\ 1=s^{3}A+s^{3}C-6s^{2}A+s^{2}B-3s^{2}C+s^{2}D+9sA-6sB+9B\end{array}$

Comparing both sides and solving for the values of A,B,C,D:

$\begin{array}{l}A+C=0,-6A+B-3C+D=0,9A-6B=0,9B=1\\ A=\frac{2}{27},B=\frac{1}{9},C=\frac{-2}{27},D=\frac{1}{9}\end{array}$

Thus equation becomes,

$\frac{1}{s^2{\left(s-3\right)}^2}=\frac{2}{27}.\frac{1}{s}+\frac{1}{9}.\frac{B}{s^2}-\frac{2}{27}.\frac{1}{s-3}+\frac{1}{9}.\frac{1}{{\left(s-3\right)}^2}$

This gives us,

$\begin{array}{l}Y\left(s\right)=\frac{1}{{\left(s-3\right)}^2}+\frac{2}{27}.\frac{1}{s}+\frac{1}{9}.\frac{B}{s^2}-\frac{2}{27}.\frac{1}{s-3}+\frac{1}{9}.\frac{1}{{\left(s-3\right)}^2}\\ Y\left(s\right)=\frac{2}{27}.\frac{1}{s}+\frac{1}{9}.\frac{B}{s^2}-\frac{2}{27}.\frac{1}{s-3}+\frac{10}{9}.\frac{1}{{\left(s-3\right)}^2}\end{array}$

Now we apply 7.3.1 and 7.2.1 theorem, to obtain the inverse Laplace transform of the given function.

$L^{-1}\left\{Y\left(s\right)\right\}=\frac{2}{27}.L^{-1}\left\{\frac{1}{s}\right\}+\frac{1}{9}.L^{-1}\left\{\frac{1}{s^2}\right\}-\frac{2}{27}.L^{-1}\left\{\frac{1}{s-3}\right\}+\frac{10}{9}.L^{-1}\left\{\frac{1}{{\left(s-3\right)}^2}\right\}$

Here, $\frac{1}{\left(s-3\right)}$is,$F\left(s\right)=\frac{1}{s}$ shifted three units to the right.

And, $\frac{1}{{\left(s-3\right)}^2}$is,data-custom-editor="chemistry" $F\left(s\right)=\frac{1}{s^2}$ shifted three units to the right.

$\begin{array}{l}{L}^{-1}\left\{Y\left(s\right)\right\}=\frac{2}{27}.L^{-1}\left\{\frac{1}{s}\right\}+\frac{1}{9}.L^{-1}\left\{\frac{1}{s^2}\right\}-\frac{2}{27}.L^{-1}\left\{\frac{1}{s-3}\right\}+\frac{10}{9}.L^{-1}\left\{\frac{1}{{\left(s-3\right)}^2}\right\}\\ y\left(t\right)=\frac{2}{27}+\frac{1}{9}t-\frac{2}{27}{e}^{3t}+\frac{10}{9}t{e}^{3t}\end{array}$