Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

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

$c_{1} = \frac{5}{3} and c_{2} = \frac{1}{3}$
$c_{1} = \frac{2}{3 e} and c_{2} = \frac{1}{3 e^{- 2}}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Taking the given function as y

First of all, take the given function as, $ϕ (x) = y$ .

Step 2: Differentiating the given function concerning x

Differentiating $ϕ (x) = y = c_{1} e^{x} + c_{2} e^{- 2 x}$ , concerning x,

$\frac{dy}{dx} = c_{1} e^{x} - 2 c_{2} e^{- 2 x}$

Again, differentiating concerning x,

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

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

$\frac{d^{2} y}{{dx}^{2}} = c_{1} e^{x} + 4 c_{2} e^{- 2 x} \frac{d^{2} y}{{dx}^{2}} = c_{1} e^{x} + 4 c_{2} e^{- 2 x} - c_{1} e^{x} + 2 c_{2} e^{- 2 x} + c_{1} e^{x} - 2 c_{2} e^{- 2 x} \begin{array}{l} \frac{d^{2} y}{{dx}^{2}} = c_{1} e^{x} + 4 c_{2} e^{- 2 x} - (c_{1} e^{x} - 2 c_{2} e^{- 2 x}) + c_{1} e^{x} - 2 c_{2} e^{- 2 x} \\ \frac{d^{2} y}{{dx}^{2}} = c_{1} e^{x} + 4 c_{2} e^{- 2 x} - \frac{dy}{dx} + c_{1} e^{x} - 2 c_{2} e^{- 2 x} \\ \frac{d^{2} y}{{dx}^{2}} = 2 c_{1} e^{x} + 2 c_{2} e^{- 2 x} - \frac{dy}{dx} \\ \frac{d^{2} y}{{dx}^{2}} = 2 (c_{1} e^{x} + c_{2} e^{- 2 x}) - \frac{dy}{dx} \\ \frac{d^{2} y}{{dx}^{2}} = 2 y - \frac{dy}{dx} \\ \frac{d^{2} y}{{dx}^{2}} = 2 y - \frac{dy}{dx} \\ \frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} - 2 y = 0 \end{array}$

Which is identical to the given differential equation.

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

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 $ϕ (x)$ ,

$\begin{array}{l} c_{1} e^{0} + c_{2} e^{0} = 2 \\ c_{1} + c_{2} = 2 \cdot \cdot \cdot \cdot \cdot \cdot (1) \end{array}$

Also, when $x = 0, y' = 1$

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

$\begin{array}{l} c_{1} e^{0} - 2 c_{2} e^{0} = 1 \\ c_{1} - 2 c_{2} = 1 \cdot \cdot \cdot \cdot \cdot \cdot (2) \end{array}$

Now, Subtracting (2) from (1),

$\begin{array}{l} 3 c_{2} = 1 \\ c_{2} = \frac{1}{3} \end{array}$

Putting this value of $c_{2}$ in (1),

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

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

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 $ϕ (x)$ ,

$c_{1} e + c_{2} e^{- 2} = 1 \cdot \cdot \cdot \cdot \cdot \cdot (3)$

Also, when $x = 1, y' = 0$

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

$c_{1} e - 2 c_{2} e^{- 2} = 0 \cdot \cdot \cdot \cdot \cdot \cdot (4)$

Now, Subtracting (4) from (3),

$3 c_{2} e^{- 2} = 1 c_{2} = \frac{1}{3 e^{- 2}}$

Putting this value of $c_{2}$ in (3),

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

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: Taking the given function as y

Step 2: Differentiating the given function concerning x

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

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

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

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Verify that the function ϕ(x)=c1ex+c2e-2x is a solution to the linear equation d2ydx2+dydx-2y=0 for any choice of the constants c1 and c2. Determine c1 and c2 so that each of the following initial conditions is satisfied.(a) y(0)=2, y'(0)=1(b) y(1)=1, y'(1)=0

Step by Step Solution

Step 1: Taking the given function as y

Step 2: Differentiating the given function concerning x

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

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

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

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