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Q25E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 172

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

Solve the given initial value problem. $y'' - 2 y' + 2 y = 0;$ $y (π) = e^{π},$ $y' (π) = 0$

The solution of the given initial value $y'' - 2 y' + 2 y = 0$ is $y (t) = e^{t} (- cost + sint)$ when $y (π) = e^{π}$ and $y' (π) = 0$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Complex conjugate roots.

If the auxiliary equation has complex conjugate roots $α \pm iβ$ , then the general solution is given as: $y (t) = c_{1} e^{αt} cosβt + c_{2} e^{αt} sinβt$

Step 2: Finding the roots of the auxiliary equation

Given differential equation is $y'' - 2 y' + 2 y = 0 .$

Then the auxiliary equation is $r^{2} - 2 r + 2 = 0 .$

$\begin{array}{l} r = \frac{2 \pm \sqrt{2^{2} - 4 \times 1 \times 2}}{2 \times 1} \\ r = \frac{2 \pm \sqrt{4 - 8}}{} \\ r = \frac{2 \pm \sqrt{- 4}}{} \\ r = \frac{2 \pm 2 i}{} \\ r = 1 \pm i \end{array}$

Therefore, the general solution is:

$\begin{matrix} y (t) = e^{1 \times t} (c_{1} \cos (t) + c_{2} \sin (t)) \\ = e^{t} (c_{1} \cos (t) + c_{2} \sin (t)) \end{matrix}$

Step 3: Finding the values of c1 and c2

Given initial conditions are $y (π) = e^{π}$ and $y' (π) = 0$

$\begin{array}{l} y (π) = e^{π} (c_{1} \cos (π) + c_{2} \sin (π)) \\ - c_{1} e^{π} = e^{π} \\ c_{1} = - 1 \end{array}$

And $y' (t) = e^{t} (c_{1} cost + c_{2} sint) + e^{t} (- c_{1} sint + c_{2} cost)$

Then $y' (π) = e^{π} (c_{1} \cos (π) + c_{2} \sin (π)) + e^{π} (- c_{1} \sin (π) + c_{2} \cos (π))$

$\begin{matrix} - e^{π} (c_{1} + c_{2}) = 0 \\ c_{1} + c_{2} = 0 \\ c_{2} = - c_{1} \end{matrix}$

Then $c_{2} = 1$

Therefore, the solution is $y (t) = e^{t} (- cost + sint)$

Most popular questions for Math Textbooks

Discontinuous Forcing Term. In certain physical models, the nonhomogeneous term, or forcing term, g(t) in the equation

$ay'' + by' + cy = g (t)$

may not be continuous but may have a jump discontinuity. If this occurs, we can still obtain a reasonable solution using the following procedure. Consider the initial value problem;

$y'' + 2 y' + 5 y = g (t); y (0) = 0, y' (0) = 0$

Where,

$g (t) = \{\begin{array}{l} 10, if 0 \leq t \leq \frac{3 π}{2} \\ 0, if t > \frac{3 π}{2} \end{array}\}$

Find a solution to the initial value problem for $0 \leq t \leq \frac{3 π}{2}$ .
Find a general solution for $t > \frac{3 π}{2}$ .
Now choose the constants in the general solution from part (b) so that the solution from part (a) and the solution from part (b) agree, together with their first derivatives, at $t = \frac{3 π}{2}$ . This gives us a continuously differentiable function that satisfies the differential equation except at $t = \frac{3 π}{2}$ .

Find a general solution $z " + 10 z' + 25 z = 0$

Find a general solution $y''' + y'' + 3 y' - 5 y = 0$

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

Solve the given initial value problem $y'' + 2 y' + 17 y = 0;$ $y (0) = 1,$ $y' (0) = - 1$ .

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