Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Solve the differential equation $\frac{d^{2} x}{{dt}^{2}} + 2 x = \cos (t)$ and find all the real solutions of the differential equation.

The solution is $x (t) = c_{1} \cos (\sqrt{2} t) + c_{2} e^{t} \sin (\sqrt{2} t) + \cos (t)$ .

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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 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.

$T (f) = x " + 2 x$

Then the characteristic polynomial is as follows.

$P_{T} (λ) = λ^{2} + 2$

Solve the characteristic polynomial and find the roots as follows.

$λ^{2} + 2 = 0 λ^{2} = - 2 λ = \pm \sqrt{2 i}$

Therefore, the roots of the characteristic polynomials are $\sqrt{2 i}$ and $- \sqrt{2 i}$ .

Step3: Simplification of the solution

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

The exponential functions $e^{\sqrt{2 it}}$ and $e^{- \sqrt{2 it}}$ 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 .

The complementary function as follows.

$f_{c} (t) = c_{1} \cos (\sqrt{2 t}) + c_{2} e^{t} \sin (\sqrt{2 t})$

Step4: To find the particular solution

The particular solution to the differential equation $x " + 2 x + = \cos (t)$ is of the form role="math" localid="1660806105789" $x_{p} (t) = Acos (t) + Bsin (t)$ .

Differentiate the function $x_{p} (t) = Acos (t) + Bsin (t)$ with respect to t as follows.

$x_{p} (t) = A \cos (t) + B \sin (t) x' = - A \sin + B \cos t$

Similarly, differentiate the function $x' = - A \sin t + B \cos t$ with respect to t as follows.

$x' = - A \sin t + B \cos t x " = - A \cos t - B \sin t$

Substitute the values $- A \cos t - B \sin t$ for x" and $A \cos (t) + B \sin (t)$ for x in x"+2x as follows.

$x " + 2 x = - Acost - Bsint + 2 (Acost + Bsint) = - Acost - Bsint + 2 Acost + 2 Bsint = Acost + Bsint x " + 2 x = Acost + Bsint$

Substitute the value $A \cos t + B \sin t$ for $x " + 2 x$ in x"+2x = cos(t) as follows.

$x " + 2 x = \cos (t) Acost + Bsint = \cos (t)$

Compare the coefficients of Cost and sint as follows.

A = 1

B = 0

Substitute the value 1 for A and 0 for B in role="math" localid="1660806778527" $x_{p} (t) = Acos (t) + Bsin (t)$ as follows.

$x_{p} (t) = Acos (t) + Bsin (t) x_{p} (t) = 1 . \cos (t) + 0 . \sin (t) x_{p} (t) = \cos (t)$

Therefore, the general solution as follows.

$x (t) = x_{c} (t) + x_{p} (t) x (t) = c_{1} \cos (\sqrt{2} t) + c_{2} e^{t} \sin (\sqrt{2} t) + \cos (t)$

Thus, the general solution of the differential equation $x " + 2 x = \cos (t)$ is $x (t) = c_{1} \cos (\sqrt{2} t) + c_{2} e^{t} \sin (\sqrt{2 t}) + \cos (t)$ .

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Linear Algebra With Applications

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Short Answer

Solve the differential equation $\frac{d^{2} x}{{dt}^{2}} + 2 x = \cos (t)$ and find all the real solutions of the differential equation.

Step by Step Solution

Step1: Definition of characteristic polynomial

Step2: Determination of the solution

Step3: Simplification of the solution

Step4: To find the particular solution

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Solve the differential equation d2xdt2+2x=cos(t) and find all the real solutions of the differential equation.

Step by Step Solution

Step1: Definition of characteristic polynomial

Step2: Determination of the solution

Step3: Simplification of the solution

Step4: To find the particular solution

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Solve the differential equation $\frac{d^{2} x}{{dt}^{2}} + 2 x = \cos (t)$ and find all the real solutions of the differential equation.