Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Solve the differential equation $f''' (t) - f'' (t) - 4 f' (t) + 4 f (t) = 0$ and find all the real solution of the differential equation.

The solution is $f (t) = C_{1} e^{t} + C_{2} e^{2 t} + C_{3} e^{- 2 t}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1: Definition of characteristic polynomial

Consider the linear differential operator

$T (f) = f^{(n)} + a_{n - 1} f^{(n - 1)} + . . . + a_{1} f' + a_{0} f$ .

The characteristic polynomial of T is defined as

$p_{T} (λ) = λ^{n} + a_{n - 1} + . . . + a_{1} λ + a_{0}$ .

Step2: Determination of the solution

The characteristic polynomial of the operator as follows.

$T (f) = f''' ‐ f'' ‐ 4 f' + 4 f$

Then the characteristic polynomial is as follows.

role="math" localid="1660798972108" $P_{T} (λ) = λ^{3} ‐ λ^{2} ‐ 4 λ + 4$

Solve the characteristic polynomial and find the roots as follows.

$λ^{3} ‐ λ^{2} ‐ 4 λ + 4 = 0 λ^{2} (λ ‐ 1) ‐ 4 (λ ‐ 1) = 0 (λ^{2} ‐ 4) (λ ‐ 1) = 0 (λ + 2) (λ ‐ 2) (λ ‐ 1) = 0$

Simplify further as follows.

$λ ‐ 1 = 0 λ = 1 λ + 2 = 0 λ = ‐ 2$

Simplify further as follows.

$λ ‐ 2 = 0 λ = 2$

Therefore, the roots of the characteristic polynomials are 1,2 and $‐ 2$ .

Step3: Conclusion of the solution

Since, the roots of the characteristic equation is different real numbers.

The exponential functions $e^{t}, e^{2 t}$ and $e^{- 2 t}$ form a basis of the kernel of T .

Hence, they form a basis of the solution space of the homogenous differential equation is $T (f) = 0$ .

Thus, the general solution of the differential equation $f''' (t) ‐ f'' (t) ‐ 4 f' (t) + 4 f' (t) = 0$ is $f (t) = C_{1} e^{t} + C_{2} e^{2 t} + C_{3} e^{- 2 t}$ .

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Linear Algebra With Applications

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Short Answer

Solve the differential equation $f''' (t) - f'' (t) - 4 f' (t) + 4 f (t) = 0$ and find all the real solution of the differential equation.

Step by Step Solution

Step1: Definition of characteristic polynomial

Step2: Determination of the solution

Step3: Conclusion of the solution

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Linear Algebra With Applications

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Short Answer

Solve the differential equation f'''(t)-f''(t)-4f'(t)+4f(t)=0and find all the real solution of the differential equation.

Step by Step Solution

Step1: Definition of characteristic polynomial

Step2: Determination of the solution

Step3: Conclusion of the solution

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Solve the differential equation $f''' (t) - f'' (t) - 4 f' (t) + 4 f (t) = 0$ and find all the real solution of the differential equation.