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### 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

# Verify that the function ${\mathbf{\varphi }}\left(x\right){\mathbf{=}}{{\mathbf{c}}}_{{\mathbf{1}}}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{+}}{{\mathbf{c}}}_{{\mathbf{2}}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$ is a solution to the linear equation $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{-}}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$ for any choice of the constants ${{\mathbf{c}}}_{{\mathbf{1}}}$ and ${{\mathbf{c}}}_{{\mathbf{2}}}$. Determine ${{\mathbf{c}}}_{{\mathbf{1}}}$ and ${{\mathbf{c}}}_{{\mathbf{2}}}$ so that each of the following initial conditions is satisfied.(a) ${\mathbf{y}}\left(0\right){\mathbf{=}}{\mathbf{2}}{\mathbf{,}}{\mathbf{}}{\mathbf{y}}{\mathbf{\text{'}}}\left(0\right){\mathbf{=}}{\mathbf{1}}$(b) ${\mathbf{y}}\left(1\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{\mathbf{}}{\mathbf{y}}{\mathbf{\text{'}}}\left(1\right){\mathbf{=}}{\mathbf{0}}$

1. ${\mathrm{c}}_{1}=\frac{5}{3}\mathrm{and}{\mathrm{c}}_{2}=\frac{1}{3}$
2. ${\mathrm{c}}_{1}=\frac{2}{3\mathrm{e}}\mathrm{and}{\mathrm{c}}_{2}=\frac{1}{3{\mathrm{e}}^{-2}}$
See the step by step solution

## Step 1: Taking the given function as y

First of all, take the given function as, $\varphi \left(\mathrm{x}\right)=\mathrm{y}$.

## Step 2: Differentiating the given function concerning x

Differentiating $\varphi \left(\mathrm{x}\right)=\mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}$, concerning x,

$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}$

Again, differentiating concerning x,

$\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}$

## Step 3: Simplification of the differential equation obtained in step 2

$\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}-{\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+2{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}+{\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}\phantom{\rule{0ex}{0ex}}\begin{array}{l}\frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}-\left({\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}\right)+{\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+4{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}-\frac{\mathrm{dy}}{\mathrm{dx}}+{\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}-2{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=2{\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+2{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=2\left({\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-2\mathrm{x}}\right)-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=2\mathrm{y}-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}=2\mathrm{y}-\frac{\mathrm{dy}}{\mathrm{dx}}\\ \frac{{\mathrm{d}}^{2}\mathrm{y}}{{\mathrm{dx}}^{2}}+\frac{\mathrm{dy}}{\mathrm{dx}}-2\mathrm{y}=0\end{array}$

Which is identical to the given differential equation.

Hence, ${\varphi }\left(\mathbf{x}\right){\mathbf{=}}{{\mathbf{c}}}_{{\mathbf{1}}}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{+}}{{\mathbf{c}}}_{{\mathbf{2}}}{{\mathbf{e}}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$ is a solution to $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}{\mathbf{+}}\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{-}}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}$, for any choice of the constants ${{\mathbf{c}}}_{{1}}$ and ${{\mathbf{c}}}_{{2}}$.

## Step 4(a): Determining c1 and satisfying the initial condition given in part (a)

As given in part (a) of the question that when x = 0, y = 2

Therefore, we will put these values in the given function $\mathrm{\varphi }\left(\mathrm{x}\right)$,

$\begin{array}{l}{\mathrm{c}}_{1}{\mathrm{e}}^{0}+{\mathrm{c}}_{2}{\mathrm{e}}^{0}=2\\ {\mathrm{c}}_{1}+{\mathrm{c}}_{2}=2······\left(1\right)\end{array}$

Also, when $\mathrm{x}=0,\mathrm{y}\text{'}=1$

Therefore, we will put these values in $\frac{\mathrm{dy}}{\mathrm{dx}}$,

$\begin{array}{l}{\mathrm{c}}_{1}{\mathrm{e}}^{0}-2{\mathrm{c}}_{2}{\mathrm{e}}^{0}=1\\ {\mathrm{c}}_{1}-2{\mathrm{c}}_{2}=1······\left(2\right)\end{array}$

Now, Subtracting (2) from (1),

$\begin{array}{l}3{\mathrm{c}}_{2}=1\\ {\mathrm{c}}_{2}=\frac{1}{3}\end{array}$

Putting this value of ${\mathrm{c}}_{2}$ in (1),

$\begin{array}{l}{\mathrm{c}}_{1}+\frac{1}{3}=2\\ {\mathrm{c}}_{1}=2-\frac{1}{3}\\ {\mathrm{c}}_{1}=\frac{6-1}{3}\\ {\mathrm{c}}_{1}=\frac{5}{3}\end{array}$

Thus, to satisfy the initial condition given in part (a), ${{\mathbf{c}}}_{{\mathbf{1}}}{\mathbf{=}}\frac{\mathbf{5}}{\mathbf{3}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbf{c}}}_{{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{3}}$.

## Step 5(b): Determining c1 and satisfying the initial condition given in part (b)

As given in part (b) of the question that when x = 1, y = 1

So, we will put these values in the given function $\mathrm{\varphi }\left(\mathrm{x}\right)$,

${\mathrm{c}}_{1}\mathrm{e}+{\mathrm{c}}_{2}{\mathrm{e}}^{-2}=1······\left(3\right)$

Also, when $\mathrm{x}=1,\mathrm{y}\text{'}=0$

Consequently, we will put these values in $\frac{\mathrm{dy}}{\mathrm{dx}}$,

${\mathrm{c}}_{1}\mathrm{e}-2{\mathrm{c}}_{2}{\mathrm{e}}^{-2}=0······\left(4\right)$

Now, Subtracting (4) from (3),

$3{\mathrm{c}}_{2}{\mathrm{e}}^{-2}=1\phantom{\rule{0ex}{0ex}}{\mathrm{c}}_{2}=\frac{1}{3{\mathrm{e}}^{-2}}$

Putting this value of ${\mathrm{c}}_{2}$ in (3),

$\begin{array}{l}{\mathrm{c}}_{1}\mathrm{e}+\frac{1}{3{\mathrm{e}}^{-2}}{\mathrm{e}}^{-2}=1\\ {\mathrm{c}}_{1}\mathrm{e}+\frac{1}{3}=1\\ {\mathrm{c}}_{1}\mathrm{e}=1-\frac{1}{3}\\ {\mathrm{c}}_{1}\mathrm{e}=\frac{2}{3}\\ {\mathrm{c}}_{1}=\frac{2}{3\mathrm{e}}\end{array}$

Accordingly, to satisfy the initial condition given in part (b), ${{\mathbf{c}}}_{{\mathbf{1}}}{\mathbf{=}}\frac{\mathbf{2}}{\mathbf{3}\mathbf{e}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{{\mathbf{c}}}_{{\mathbf{2}}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{2}}}$.