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Expert-verified Found in: Page 413 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # use the method of Laplace transforms to solve the given initial value problem. Here x′, y′, etc., denotes differentiation with respect to t; so does the symbol D. ${{\mathbit{x}}}^{{\mathbf{\text{'}}}}{\mathbf{-}}{\mathbf{2}}{\mathbit{y}}{\mathbf{=}}{\mathbf{2}}$${\mathbit{x}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$ ${{\mathbit{x}}}^{{\mathbf{\text{'}}}}{\mathbf{+}}{\mathbit{x}}{\mathbf{-}}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{=}}{{\mathbit{t}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{2}}{\mathbit{t}}{\mathbf{-}}{\mathbf{1}}$${\mathbit{y}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}$

The solution is $x\left(t\right)={t}^{2},y\left(t\right)=t-1$.

See the step by step solution

## Step 1: Given information

Since the initial conditions are given at $t=1$, we need to shift the argument to the zero, so let

$\begin{array}{l}{x}_{1}\left(t\right)=x\left(t+1\right)\\ {y}_{1}\left(t\right)=y\left(t+1\right)\end{array}$

Then ${x}_{1}\left(0\right)=x\left(1\right)=1$ and ${y}_{1}\left(0\right)=y\left(1\right)=0$ and

$\begin{array}{l}{x}_{1}^{\text{'}}\left(t\right)={x}^{\text{'}}\left(t+1\right)\left(t+1\right)={x}^{\text{'}}\left(t+1\right)\\ {y}_{1}^{\text{'}}\left(t\right)={y}^{\text{'}}\left(t+1\right)\left(t+1\right)={y}^{\text{'}}\left(t+1\right)\end{array}$

Then the given system becomes

$\left\{\begin{array}{l}{x}_{1}^{\text{'}}-2{y}_{1}=2\\ {x}_{1}^{\text{'}}+{x}_{1}-{y}_{1}^{\text{'}}={\left(t+1\right)}^{2}-2\left(t+1\right)-1\end{array}⇒\left\{\begin{array}{l}{x}_{1}^{\text{'}}-2{y}_{1}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{x}_{1}\left(0\right)=1\\ {x}_{1}^{\text{'}}+{x}_{1}-{y}_{1}^{\text{'}}={t}^{2}+4t+2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{y}_{1}\left(0\right)=0\end{array}$

## Step 2: Take Laplace transform

Applying the Laplace transform gives

$\left\{\begin{array}{l}L\left\{{x}_{1}^{\text{'}}-2{y}_{1}\right\}\left(s\right)=L\left\{2\right\}\left(s\right)\\ L\left\{{x}_{1}^{\text{'}}+{x}_{1}-{y}_{1}^{\text{'}}\right\}\left(s\right)=L\left\{{t}^{2}+4t+2\right\}\left(s\right)\end{array}$

Simplify equation , further as:

$⇒\left\{\begin{array}{l}s{X}_{1}\left(s\right)-{x}_{1}\left(0\right)-2{Y}_{1}\left(s\right)=\frac{2}{s}\\ s{X}_{1}\left(s\right)-{x}_{1}\left(0\right)+{X}_{1}\left(s\right)-s{Y}_{1}\left(s\right)-{y}_{1}\left(0\right)=\frac{2}{{s}^{3}}+\frac{4}{{s}^{2}}+\frac{2}{s}\end{array}$

$⇒\left\{\begin{array}{l}s{X}_{1}\left(s\right)-1+2{Y}_{1}\left(s\right)=\frac{2}{s}\\ s{X}_{1}\left(s\right)-1+{X}_{1}\left(s\right)-s{Y}_{1}\left(s\right)=\frac{2}{{s}^{3}}+\frac{4}{{s}^{2}}+\frac{2}{s}\end{array}$

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

Simplify equation$\left(2+\frac{2}{s}-s\right){Y}_{1}\left(s\right)=\frac{2}{{s}^{3}}-\frac{1}{s}+\frac{2}{{s}^{2}}$ , further as:

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

## Step 3: Finding the functions

Substituting $\frac{1}{{s}^{2}}$ for ${Y}_{1}\left(s\right)$ in equation (1) gives

${X}_{1}\left(s\right)=\frac{2}{{s}^{2}}+\frac{2}{{s}^{3}}+\frac{1}{s}$

Applying inverse Laplace we get

$\begin{array}{l}{L}^{-1}\left\{{Y}_{1}\left(s\right)\right\}\left(t\right)={L}^{-1}\left\{\frac{1}{{s}^{2}}\right\}\left(t\right)Þ{y}_{1}\left(t\right)=t\\ {L}^{-1}\left\{{X}_{1}\left(s\right)\right\}\left(t\right)={L}^{-1}\left\{\frac{2}{{s}^{2}}+\frac{2}{{s}^{3}}+\frac{1}{s}\right\}\left(t\right)\end{array}$

Take inverse Laplace transform we get;

$\begin{array}{c}{x}_{1}\left(t\right)={t}^{2}+2t+1\\ ={\left(t+1\right)}^{2}\end{array}$

Now we find

$\begin{array}{l}x\left(t\right)={x}_{1}\left(t-1\right)={t}^{2}\\ y\left(t\right)={y}_{1}\left(t-1\right)=t-1\end{array}$

Therefore

$x\left(t\right)={t}^{2},y\left(t\right)=t-1$ ### Want to see more solutions like these? 