Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{=}{\mathit{t}}^{\mathbf{3}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}$

Short answer

The solution of above initial-value problem is,

$y\left(t\right)=\frac{3}{4}+\frac{9}{8}t+\frac{3}{4}{t}^2+\frac{1}{4}{t}^3+\frac{1}{4}{e}^{2t}-\frac{13}{8}t{e}^{2t}$

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 $\mu \left(t-a\right)$is,

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

Step 2:

TakingLaplace transform of each member of differential equation:

$\begin{array}{l}y\text{'}\text{'}-4y\text{'}+4y=t^3,y\left(0\right)=1,y\text{'}\left(0\right)=0\\ L\left\{y\text{'}\text{'}\right\}-4L\left\{y\text{'}\right\}+4L\left\{y\right\}=L\left\{t^3\right\}\end{array}$

We get;

$s^{2}Y\left(s\right)-sy\left(0\right)-y\text{'}\left(0\right)-4sY\left(s\right)+4y\left(0\right)+4Y\left(s\right)=\frac{3!}{s^4}$

Rearranging the term and writing the equation in terms of $Y\left(s\right)$and using the initial value to get.

$\begin{array}{c}{s}^{2}Y\left(s\right)-s-4sY\left(s\right)+4+4Y\left(s\right)=\frac{6}{s^4}\\ \left(s^2-4s+4\right)Y\left(s\right)=s-4+\frac{6}{s^2}\\ Y\left(s\right)=\frac{s-4}{{\left(s-2\right)}^2}+\frac{6}{s^4{\left(s-2\right)}^2}\end{array}$

Step 3:

$\begin{array}{c}{L}^{-1}\left\{Y\left(s\right)\right\}=L^{-1}\left\{\frac{s-4}{{\left(s-2\right)}^2}\right\}+L^{-1}\left\{\frac{6}{s^4{\left(s-2\right)}^2}\right\}\\ \frac{6}{s^4{\left(s-2\right)}^2}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s^3}+\frac{D}{s^4}+\frac{E}{s-2}+\frac{F}{{\left(s-2\right)}^2}\\ \frac{1}{s^4{\left(s-2\right)}^2}=\frac{s^3{\left(s-2\right)}^{2}A+B{s}^2{\left(s-2\right)}^2+s{\left(s-2\right)}^{2}C+D{\left(s-2\right)}^2+s^4\left(s-2\right)E+s^{4}F}{s^4{\left(s-2\right)}^2}\end{array}$

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

Equating numerators, we get

$\begin{array}{l}1=s^3{\left(s-2\right)}^{2}A+B{s}^2{\left(s-2\right)}^2+s{\left(s-2\right)}^{2}C+D{\left(s-2\right)}^2+s^4\left(s-2\right)E+s^{4}F\\ 6=s^5\left(A+E\right)+s^4\left(-4A+B-2E+F\right)+s^3\left(4A-4B+C\right)+s^2\left(4B-4C+D\right)+s\left(4C-4D\right)+4D\end{array}$

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

$\begin{array}{l}A+E=0,-4A+B-2E+F=0,4A-4B+C=0,4B-4C+D=0,4C-4D=0,4D=6\\ A=\frac{3}{4},B=\frac{9}{8},C=D=\frac{3}{2},E=-\frac{3}{4},F=\frac{3}{8}\end{array}$

Thus equation becomes,

$\frac{6}{s^4{\left(s-2\right)}^2}=\frac{3}{4}\frac{1}{s}+\frac{9}{8}\frac{1}{s^2}+\frac{3}{2}\frac{1}{s^3}+\frac{3}{2}\frac{1}{s^4}-\frac{3}{4}\frac{1}{s-2}+\frac{3}{8}\frac{1}{{\left(s-2\right)}^2}$

This gives us,

$\begin{array}{l}Y\left(s\right)=\frac{\left(s-2\right)-2}{{\left(s-2\right)}^2}+\frac{3}{4}\frac{1}{s}+\frac{9}{8}\frac{1}{s^2}+\frac{3}{2}\frac{1}{s^3}+\frac{3}{2}\frac{1}{s^4}-\frac{3}{4}\frac{1}{s-2}+\frac{3}{8}\frac{1}{{\left(s-2\right)}^2}\\ Y\left(s\right)=\frac{1}{\left(s-2\right)}-2\left\{\frac{1}{{\left(s-2\right)}^2}\right\}+\frac{3}{4}\frac{1}{s}+\frac{9}{8}\frac{1}{s^2}+\frac{3}{2}\frac{1}{s^3}+\frac{3}{2}\frac{1}{s^4}-\frac{3}{4}\frac{1}{s-2}+\frac{3}{8}\frac{1}{{\left(s-2\right)}^2}\\ Y\left(s\right)=\frac{3}{4}\left\{\frac{1}{s}\right\}+\frac{9}{8}\left\{\frac{1}{s^2}\right\}+\frac{3}{2}\left\{\frac{1}{s^3}\right\}+\frac{3}{2}\left\{\frac{1}{s^4}\right\}+\frac{1}{4}\left\{\frac{1}{s-2}\right\}-\frac{13}{8}\left\{\frac{1}{{\left(s-2\right)}^2}\right\}\end{array}$

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

Here,$\frac{1}{\left(s-2\right)}$ is, $F\left(s\right)=\frac{1}{s}$shifted two units to the right.

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

$L^{-1}\left\{Y\left(s\right)\right\}=\frac{3}{4}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{9}{8}{L}^{-1}\left\{\frac{1}{s^2}\right\}+\frac{3}{2}{L}^{-1}\left\{\frac{1}{s^3}\right\}+\frac{3}{2}{L}^{-1}\left\{\frac{1}{s^4}\right\}+\frac{1}{4}{L}^{-1}\left\{\frac{1}{s-2}\right\}-\frac{13}{8}{L}^{-1}\left\{\frac{1}{{\left(s-2\right)}^2}\right\}$

Multiplying and dividing the third and forth term by 2! And 3!, respectively

$\begin{array}{l}{L}^{-1}\left\{Y\left(s\right)\right\}=\frac{3}{4}{L}^{-1}\left\{\frac{1}{s}\right\}+\frac{9}{8}{L}^{-1}\left\{\frac{1}{s^2}\right\}+\frac{3}{2\left(2!\right)}{L}^{-1}\left\{\frac{2!}{s^3}\right\}+\frac{3}{2\left(3!\right)}{L}^{-1}\left\{\frac{3!}{s^4}\right\}+\frac{1}{4}{L}^{-1}\left\{{\left.\frac{1}{s}\right|}_{s\to s-2}\right\}-\frac{13}{8}{L}^{-1}\left\{{\left.\frac{1}{s^2}\right|}_{s\to s-2}\right\}\\ y\left(t\right)=\frac{3}{4}+\frac{9}{8}t+\frac{3}{4}{t}^2+\frac{1}{4}{t}^3+\frac{1}{4}{e}^{2t}-\frac{13}{8}t{e}^{2t}\end{array}$