Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Solve the initial value problem in $f'' (t) ‐ 9 f (t) = 0; f (0) = 0, f' (0) = 1$

The solution is $f (t) = \cos (3 t)$ .

See the step by step solution

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 is defined as

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

Step2: Determination of the solution

The characteristic polynomial of the operator as follows.

$T (f) = f' - 9$

Then the characteristic polynomial is as follows.

$P_{T} (λ) = λ^{2} - 9$

Solve the characteristic polynomial and find the roots as follows.

$λ^{2} - 9 = 0 λ^{2} = 9 λ = \sqrt{9} λ = \pm 3$

Therefore, the roots of the characteristic polynomials are 3 and -3.

Step3: Explanation of the solution

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

The exponential functions $e^{3 t} and e^{- 3 t}$ 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 $f' (t) - 9 f' (t) = 0 is f (t) = c_{1} e^{3 t} + c_{2} e^{- 3 t}$ .

Step 4: Explanation of the solution with initial value problem

Consider the general solution of the $f' " (t) - 9 f " (t) = 0$ with the initial value problem $f (0) = 0, f' (0) = 1$ is as follows

role="math" localid="1660809514458" $f (t) = c_{1} e^{3 t} + c_{2} e^{- 3 t} f (t) = 0 (e^{3 t}) + 1 (e^{- 3 t}) f (t) = 0 + e^{- 3 t} f (t) = e^{- 3 t}$

Simplify further.

$= \cos (3 t)$

Hence the solution is $f (t) = \cos (3 t)$ .

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Linear Algebra With Applications

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Short Answer

Solve the initial value problem in $f'' (t) ‐ 9 f (t) = 0; f (0) = 0, f' (0) = 1$

Step by Step Solution

Step1: Definition of characteristic polynomial

Step2: Determination of the solution

Step3: Explanation of the solution

Step 4: Explanation of the solution with initial value problem

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Linear Algebra With Applications

Answers without the blur.

Short Answer

Solve the initial value problem in f''(t)‐9f(t)=0;f(0)=0,f'(0)=1

Step by Step Solution

Step1: Definition of characteristic polynomial

Step2: Determination of the solution

Step3: Explanation of the solution

Step 4: Explanation of the solution with initial value problem

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Solve the initial value problem in $f'' (t) ‐ 9 f (t) = 0; f (0) = 0, f' (0) = 1$