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Q10E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 180

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 a particular solution to the differential equation.
$y'' + 2 y' - y = 10$

Thus, the particular solution is $y_{p} (x) = - 10 .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Firstly, write the auxiliary equation of the given differential equation.

Given the differential equation,

$y'' + 2 y' - y = 10 \dots (1)$

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

$y'' + 2 y' - y = 0$

The auxiliary equation for the above equation,

$m^{2} + 2 m - 1 = 0$

Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{matrix} m^{2} + 2 m - 1 = 0 \\ m = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (- 1)}}{2 (1)} \\ m = \frac{- 2 \pm \sqrt{8}}{2} \\ m = - 1 \pm \sqrt{2} \end{matrix}$

The roots of the auxiliary equation are:

$m_{1} = - 1 + \sqrt{2} & m_{2} = - 1 - \sqrt{2}$

The complementary solution of the given equation is:

$y_{c} (x) = e^{- t} [c_{1} \cosh (\sqrt{2} t) + c_{2} \sinh (\sqrt{2} t)]$

Step 3: Final conclusion, find a particular solution to the differential equation.

According to the method of undetermined coefficients, assume the particular solution of equation (1),

$y_{p} (x) = A \dots (2)$

Now find the derivative of the above equation,

$\begin{matrix} y_{p}' (x) = 0 \\ y_{p}'' (x) = 0 \end{matrix}$

From the equation (1),

$\begin{matrix} y_{p}'' + 2 y_{p}' - y_{p} = 10 \\ (0) + 2 (0) - A = 10 \\ A = - 10 \end{matrix}$

Substitute the value of A in the equation (2), and we get:

$y_{p} (x) = - 10 .$

Therefore, the particular solution of the differential equation is:

$y_{p} (x) = - 10$

Most popular questions for Math Textbooks

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$y'' = 6 t; y (0) = 3, y' (0) = - 1$

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