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Expert-verified Found in: Page 186 ### 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 # Find the solution to the initial value problem.${\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{x}\right){\mathbf{+}}{\mathbf{z}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{2}}{{\mathbf{e}}}^{-x}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{z}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{z}}{\mathbf{\text{'}}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{0}}$

The solution to the initial value problem is:

$\mathrm{z}=\mathrm{sinx}-\mathrm{cosx}+{\mathrm{e}}^{-\mathrm{x}}$.

See the step by step solution

## Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

$\mathrm{z}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{z}\left(\mathrm{x}\right)=2{\mathrm{e}}^{-\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

Write the homogeneous differential equation of the equation (1),

$\mathrm{z}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{z}\left(\mathrm{x}\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^{2}+1=0$

## Step 2: Now find the complementary solution of the given equation.

The root of an auxiliary equation is,

The complementary solution of the given equation is,

${\mathrm{z}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{c}}_{1}\mathrm{cosx}+{\mathrm{c}}_{2}\mathrm{sinx}$

## Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{Ae}}^{-\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)$

Now find the first and second derivatives of the above equation,

$\begin{array}{rcl}{\mathrm{z}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)& =& -{\mathrm{Ae}}^{-\mathrm{x}}\\ {\mathrm{z}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)& =& {\mathrm{Ae}}^{-\mathrm{x}}\end{array}$

Substitute the value of and the equation (1),

$\begin{array}{c}\mathrm{z}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{z}\left(\mathrm{x}\right)=2{\mathrm{e}}^{-\mathrm{x}}\\ {\mathrm{Ae}}^{-\mathrm{x}}+{\mathrm{Ae}}^{-\mathrm{x}}=2{\mathrm{e}}^{-\mathrm{x}}\\ 2{\mathrm{Ae}}^{-\mathrm{x}}=2{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{A}=1\end{array}$

Substitute the value of A in the equation (2),

${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{x}}$

## Step 4: Find the general solution and use the given initial condition.

Therefore, the general solution is,

$\begin{array}{l}\mathrm{z}={\mathrm{z}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{z}={\mathrm{c}}_{1}\mathrm{cosx}+{\mathrm{c}}_{2}\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(3\right)\end{array}$

Given the initial condition,

$\mathrm{z}\left(0\right)=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{z}\text{'}\left(0\right)=0$

Substitute the value of z = 0 and x = 0 in the equation (3),

$\begin{array}{c}\mathrm{z}={\mathrm{c}}_{1}\mathrm{cosx}+{\mathrm{c}}_{2}\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}\\ 0={\mathrm{c}}_{1}\mathrm{cos}\left(0\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(0\right)+{\mathrm{e}}^{-0}\\ {\mathrm{c}}_{1}=-1\end{array}$

Now find the derivative of the equation (3),

$\mathrm{z}\text{'}=-{\mathrm{c}}_{1}\mathrm{sinx}+{\mathrm{c}}_{2}\mathrm{cosx}-{\mathrm{e}}^{-\mathrm{x}}$

Substitute the value of z’ = 0 and x = 0 in the above equation,

$\begin{array}{c}\mathrm{z}\text{'}=-{\mathrm{c}}_{1}\mathrm{sinx}+{\mathrm{c}}_{2}\mathrm{cosx}-{\mathrm{e}}^{-\mathrm{x}}\\ 0=-{\mathrm{c}}_{1}\mathrm{sin}\left(0\right)+{\mathrm{c}}_{2}\mathrm{cos}\left(0\right)-{\mathrm{e}}^{-0}\\ {\mathrm{c}}_{2}=1\end{array}$

Substitute the value of ${\mathrm{c}}_{1}=-1$ and ${\mathrm{c}}_{2}=1$ in the equation (3), we get:

$\begin{array}{c}\mathrm{z}={\mathrm{c}}_{1}\mathrm{cosx}+{\mathrm{c}}_{2}\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{z}=\left(-1\right)\mathrm{cosx}+\left(1\right)\mathrm{sinx}+{\mathrm{e}}^{-\mathrm{x}}\\ \mathrm{z}=\mathrm{sinx}-\mathrm{cosx}+{\mathrm{e}}^{-\mathrm{x}}\end{array}$

Thus, the solution of the initial problem is:

$\mathrm{z}=\mathrm{sinx}-\mathrm{cosx}+{\mathrm{e}}^{-\mathrm{x}}$ ### Want to see more solutions like these? 