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Q25E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 186

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

Find the solution to the initial value problem.
$z'' (x) + z (x) = 2 e^{- x}, z (0) = 0, z' (0) = 0$

The solution to the initial value problem is:

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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

The differential equation is,

$z'' (x) + z (x) = 2 e^{- x} \dots (1)$

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

$z'' (x) + z (x) = 0$

The auxiliary equation for the above equation,

$m^{2} + 1 = 0$

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

The root of an auxiliary equation is,

$\begin{matrix} m^{2} + 1 = 0 \\ m = \pm i \end{matrix}$

The complementary solution of the given equation is,

$z_{c} (x) = c_{1} cosx + c_{2} sinx$

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

Assume, the particular solution of equation (1),

$z_{p} (x) = {Ae}^{- x} . ..... (2)$

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

$\begin{array}{rcl} z_{p}' (x) & = & - {Ae}^{- x} \\ z_{p}'' (x) & = & {Ae}^{- x} \end{array}$

Substitute the value of and the equation (1),

$\begin{matrix} z'' (x) + z (x) = 2 e^{- x} \\ {Ae}^{- x} + {Ae}^{- x} = 2 e^{- x} \\ 2 {Ae}^{- x} = 2 e^{- x} \\ A = 1 \end{matrix}$

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

$z_{p} (x) = e^{- x}$

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

Therefore, the general solution is,

$\begin{array}{l} z = z_{c} (x) + z_{p} (x) \\ z = c_{1} cosx + c_{2} sinx + e^{- x} . ..... (3) \end{array}$

Given the initial condition,

$z (0) = 0, z' (0) = 0$

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

$\begin{matrix} z = c_{1} cosx + c_{2} sinx + e^{- x} \\ 0 = c_{1} \cos (0) + c_{2} \sin (0) + e^{- 0} \\ c_{1} = - 1 \end{matrix}$

Now find the derivative of the equation (3),

$z' = - c_{1} sinx + c_{2} cosx - e^{- x}$

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

$\begin{matrix} z' = - c_{1} sinx + c_{2} cosx - e^{- x} \\ 0 = - c_{1} \sin (0) + c_{2} \cos (0) - e^{- 0} \\ c_{2} = 1 \end{matrix}$

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

$\begin{matrix} z = c_{1} cosx + c_{2} sinx + e^{- x} \\ z = (- 1) cosx + (1) sinx + e^{- x} \\ z = sinx - cosx + e^{- x} \end{matrix}$

Thus, the solution of the initial problem is:

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

Most popular questions for Math Textbooks

In Problems 33, use the method of undetermined coefficients to find a particular solution to the given higher-order equation. $y''' - y'' + y = sint$

Find a particular solution to the differential equation.

$y'' (x) + y (x) = 4 xcosx$

Find a particular solution to the differential equation.

$y'' (θ) - 7 y' (θ) = θ^{2}$

Find a general solution to the differential equation.

$y'' (x) - 3 y' (x) + 2 y (x) = e^{x} sinx$

Find a particular solution to the given higher-order equation.

$y^{4} - 5 y'' + 4 y = 10 cost - 20 sint$

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