Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{t}}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{0}$

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, $a>0$hen

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

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

role="math" localid="1664301118320" $L\left\{\mu \left(t-a\right)\right\}=\frac{e^{-as}}{s}$

Step 2:

TakingLaplace transform of each member of differential equation:

$\begin{array}{l}y\text{'}\text{'}-y=e^t\cos t,y\left(0\right)=0,y\text{'}\left(0\right)=0\\ L\left\{y\text{'}\text{'}\right\}-L\left\{y\right\}=L\left\{e^t\cos t\right\}\end{array}$

We get;

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

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

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

Step 3:

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

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

Equating numerators, we get

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

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

$\begin{array}{l}A+B=0,-2A+C=0,2A=1\\ A=\frac{1}{2},B=-\frac{1}{2},C=1\end{array}$

Thus equation becomes,

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

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

$L^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{2}.L^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}.L^{-1}\left\{\frac{s-1}{{\left(s-1\right)}^2+1}\right\}-\frac{1}{2}{L}^{-1}\left\{\frac{1}{{\left(s-1\right)}^2+1}\right\}$given function.

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

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

$\begin{array}{l}{L}^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{2}.L^{-1}\left\{\frac{1}{s}\right\}-\frac{1}{2}.L^{-1}\left\{{\left.\frac{s}{s^2+1}\right|}_{s\to s-1}\right\}+\frac{1}{2}{L}^{-1}\left\{{\left.\frac{1}{s^2+1}\right|}_{s\to s-1}\right\}\\ L^{-1}\left\{Y\left(s\right)\right\}=\frac{1}{2}-\frac{1}{2}{e}^t\cos t+\frac{1}{2}{e}^t\sin t\end{array}$