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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 # In Problems 1–19, 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.$\begin{array}{l}\mathbf{D}\mathbf{\left[}\mathbf{x}\mathbf{\right]}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\mathbf{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{7}\mathbf{/}\mathbf{4}\mathbf{,}\\ \mathbf{4}\mathbf{x}\mathbf{+}\mathbf{D}\mathbf{\left[}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathbf{3}\mathbf{;}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\end{array}$

The solution is $\begin{array}{l}x\left(t\right)=\frac{3}{4}+\frac{3{e}^{-2t}}{2}-\frac{1{e}^{2t}}{2},y\left(t\right)=3{e}^{-2t}+{e}^{2t}\\ \end{array}$

See the step by step solution

## Step 1: Given information

The differential equations are given as:

$\begin{array}{l}D\left[x\right]+y=0;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x\left(0\right)=7/4,\\ 4x+D\left[y\right]=3;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y\left(0\right)=4\end{array}$

## Step 2: Apply the Laplace transform

Given initial value equations are,

$\begin{array}{l}D\left[x\right]+y=0;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x\left(0\right)=7/4,\\ \therefore {x}^{\text{'}}+y=0....\left(1\right)\\ 4x+D\left[y\right]=3;\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y\left(0\right)=4\\ \therefore 4x+{y}^{\text{'}}=3.....\left(2\right)\end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{l}sx\left(s\right)-x\left(0\right)+y\left(s\right)=0\\ sx\left(s\right)-\frac{7}{4}+y\left(s\right)=0\\ y\left(s\right)=\frac{7}{4}-sx\left(s\right).....\left(3\right)\end{array}$

Taking Laplace transform of equation second we get

$\begin{array}{r}4x\left(s\right)+sy\left(s\right)-y\left(0\right)=\frac{3}{s}\\ 4x\left(s\right)+sy\left(s\right)-4=\frac{3}{s}...\left(4\right)\end{array}$

Putting equation third into fourth we get

$\begin{array}{l}4x\left(s\right)+s\left[\frac{7}{4}-sx\left(s\right)\right]-4=\frac{3}{s}\\ \left[4-{s}^{2}\right]x\left(s\right)=\frac{12+16s-7{s}^{2}}{4s}\\ x\left(s\right)=\frac{7{s}^{2}-16s-12}{4s\left({s}^{2}-4\right)}\end{array}$

Using partial fraction we can write as

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

Taking inverse Laplace transform we get

$x\left(t\right)=\frac{3}{4}+\frac{3{e}^{-2t}}{2}-\frac{1{e}^{2t}}{2}$

Since equation first is,

$\begin{array}{l}{x}^{\text{'}}+y=0\\ -3{e}^{-2t}-{e}^{2t}+y\left(t\right)=0\\ y\left(t\right)=3{e}^{-2t}+{e}^{2t}\end{array}$

Hence

$\begin{array}{l}x\left(t\right)=\frac{3}{4}+\frac{3{e}^{-2t}}{2}-\frac{1{e}^{2t}}{2},y\left(t\right)=3{e}^{-2t}+{e}^{2t}\\ \end{array}$

## Step 3: Conclusion

The final solution is

$\begin{array}{l}x\left(t\right)=\frac{3}{4}+\frac{3{e}^{-2t}}{2}-\frac{1{e}^{2t}}{2},y\left(t\right)=3{e}^{-2t}+{e}^{2t}\\ \end{array}$ ### Want to see more solutions like these? 