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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.$\begin{array}{lll}\mathbf{\text{x''+y=1;}}& \mathbf{\text{x(0)=1,}}& {\mathbf{\text{x}}}^{\mathbf{\text{'}}}\mathbf{\text{(0)=1}}\\ {\mathbf{\text{x+y}}}^{\mathbf{\text{''}}}\mathbf{\text{=-1;}}& \mathbf{\text{y(0)=1,}}& {\mathbf{\text{y}}}^{\mathbf{\text{'}}}\mathbf{\text{(0)=-1}}\end{array}$

The solution is

$x\left(t\right)=cost+{e}^{t}-1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y\left(t\right)=cost-{e}^{t}+1$

See the step by step solution

## Step 1: Given information

The differential equations are given as:

$\begin{array}{lll}x\text{'}\text{'}+y=1;& x\left(0\right)=1,& {x}^{\text{'}}\left(0\right)=1\\ x+{y}^{\text{'}\text{'}}=-1;& y\left(0\right)=1,& {y}^{\text{'}}\left(0\right)=-1\end{array}$

## Step 2: Apply the Laplace transform

Given initial value equations are,

$\begin{array}{lll}x\text{'}\text{'}+y=1;& x\left(0\right)=1,& {x}^{\text{'}}\left(0\right)=1....\left(\text{1}\right)\\ x+{y}^{\text{'}\text{'}}=-1;& y\left(0\right)=1,& {y}^{\text{'}}\left(0\right)=-1....\left(\text{2}\right)\end{array}$

Taking Laplace transform of equation first we get

$\begin{array}{l}{s}^{2}x\left(s\right)-sx\left(0\right)-{x}^{\text{'}}\left(0\right)+y\left(s\right)=\frac{1}{s}\\ {s}^{2}x\left(s\right)-s-1+y\left(s\right)=\frac{1}{s}\\ {s}^{2}x\left(s\right)+y\left(s\right)=\frac{1+{s}^{2}+s}{s}....\left(3\right)\end{array}$

Taking Laplace transform of equation second we get

$\begin{array}{l}x\left(s\right)+{s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\text{'}}\left(0\right)=\frac{1}{s}\\ x\left(s\right)+{s}^{2}y\left(s\right)-s+1=-\frac{1}{s}\\ x\left(s\right)+{s}^{2}y\left(s\right)=\frac{{s}^{2}-s-1}{s}.....\left(4\right)\end{array}$

Solving equation third and fourth we get

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

Using partial fraction we can write as

$y\left(s\right)=\frac{s}{{s}^{2}+1}-\frac{1}{s-1}+\frac{1}{s}$

Taking inverse Laplace transform we get

$y\left(t\right)=cost-{e}^{t}+1$

## Step 3: Solve the third and fourth equation

On solving equation (3) and (4) we, obtain

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

Using partial fraction we can write as

$x\left(s\right)=\frac{s}{{s}^{2}+1}+\frac{1}{s-1}-\frac{1}{s}$

Taking inverse Laplace transform we get

$x\left(t\right)=cost+{e}^{t}-1$

Hence

$x\left(t\right)=cost+{e}^{t}-1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y\left(t\right)=cost-{e}^{t}+1$

## Step 4: conclusion

The final solution is

$x\left(t\right)=cost+{e}^{t}-1,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}y\left(t\right)=cost-{e}^{t}+1$ ### Want to see more solutions like these? 