Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question

Kindly provide a detailed and thorough explanation about this example figure. It is only one example in two pictures.

Determine the general solution to (D² + 4)(f) = sin(2x) satisfying the initial condition: f (0) = 0, D(f(0)) = 0 Solution: The related homogeneous ODE is (D² + 4)(f) = 0 with characteristic polynomial r2 + 4 = 0. The roots are +2i (multiplicity 1). Thus, the general solution to the homogeneous solution is uc(x) = c2 cos(2x) + c1 sin(2x) Finding an annihilator for sin 2x, A(D) = (D - 2i)(D + 2i) = D² + 4. Thus, a solution to the original equation also satisfies (D² + 4)(D² + 4)f) = (D² + 4)(sin(2x)) (D² + 4)²(f) = 0 The characteristic polynomial is (r² + 4)² with root +2i (multiplicity 2). The corresponding solutions are e0(x) cos(2x) = cos(2x); e0(x) sin(2x) = sin(2x) xe0(x) cos(2x) = x cos(2x); xeº(x) sin(2x) = x sin(2x) ±2i
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Transcribed Image Text:Determine the general solution to (D² + 4)(f) = sin(2x) satisfying the initial condition: f (0) = 0, D(f(0)) = 0 Solution: The related homogeneous ODE is (D² + 4)(f) = 0 with characteristic polynomial r2 + 4 = 0. The roots are +2i (multiplicity 1). Thus, the general solution to the homogeneous solution is uc(x) = c2 cos(2x) + c1 sin(2x) Finding an annihilator for sin 2x, A(D) = (D - 2i)(D + 2i) = D² + 4. Thus, a solution to the original equation also satisfies (D² + 4)(D² + 4)f) = (D² + 4)(sin(2x)) (D² + 4)²(f) = 0 The characteristic polynomial is (r² + 4)² with root +2i (multiplicity 2). The corresponding solutions are e0(x) cos(2x) = cos(2x); e0(x) sin(2x) = sin(2x) xe0(x) cos(2x) = x cos(2x); xeº(x) sin(2x) = x sin(2x) ±2i
Hence, our particular solution is of the form u,(x) = c4x cos(2x) + c3x sin(2x) + C2 cos(2x) + c, sin(2x) which can be further simplified into up(x) = c4x cos(2x)+ c3x sin(2x) Determining the values of the constants, (D² + 4)(f) = sin(2x) (D² + 4)(c4x cos(2x) + c3x sin(2x)) = sin(2x) D²(c4x cos(2x) + c3x sin(2x)) + 4(c4x cos(2x) + c3x sin(2x)) = sin(2x) D(-2c,x sin(2x) + C4 cos(2x) + 2c3x cos(2x) + cz sin(2x)) + 4c,x cos(2x) + 4c3x sin(2x) = sin(2x) -4c4 sin(2x) + 4c3 cos(2x) = sin(2x) We see that 4c3 = 0 and -4c, = 1. Hence, %3D C4 = Therefore, u, (x) = -x cos(2x) and the complete solution is of the form 1 -x cos(2x) + c2 cos(2x) + c, sin(2x) We use the initial conditions to determine c2 and c . Using the first initial condition, f(0) = c2 cos(0) + c; sin(0) = c2 = 0 Using the second initial condition and substituting the value of c2, 1 Df D(-x cos(2x) + c2 cos(2x) + c; sin(2x)) %3D = D : cos(2x) + c, sin(2x) 1 =x sin(2x) –cos(2x) + 2c, cos(2x) 1 1 D(f(0)) = -cos(0) + 2c, cos(0) = -i+ 2c, = 0 Hence, c, = Therefore, the complete solution to the problem is 1 1 f(x) = -x cos(2x) + sin(2x)
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Transcribed Image Text:Hence, our particular solution is of the form u,(x) = c4x cos(2x) + c3x sin(2x) + C2 cos(2x) + c, sin(2x) which can be further simplified into up(x) = c4x cos(2x)+ c3x sin(2x) Determining the values of the constants, (D² + 4)(f) = sin(2x) (D² + 4)(c4x cos(2x) + c3x sin(2x)) = sin(2x) D²(c4x cos(2x) + c3x sin(2x)) + 4(c4x cos(2x) + c3x sin(2x)) = sin(2x) D(-2c,x sin(2x) + C4 cos(2x) + 2c3x cos(2x) + cz sin(2x)) + 4c,x cos(2x) + 4c3x sin(2x) = sin(2x) -4c4 sin(2x) + 4c3 cos(2x) = sin(2x) We see that 4c3 = 0 and -4c, = 1. Hence, %3D C4 = Therefore, u, (x) = -x cos(2x) and the complete solution is of the form 1 -x cos(2x) + c2 cos(2x) + c, sin(2x) We use the initial conditions to determine c2 and c . Using the first initial condition, f(0) = c2 cos(0) + c; sin(0) = c2 = 0 Using the second initial condition and substituting the value of c2, 1 Df D(-x cos(2x) + c2 cos(2x) + c; sin(2x)) %3D = D : cos(2x) + c, sin(2x) 1 =x sin(2x) –cos(2x) + 2c, cos(2x) 1 1 D(f(0)) = -cos(0) + 2c, cos(0) = -i+ 2c, = 0 Hence, c, = Therefore, the complete solution to the problem is 1 1 f(x) = -x cos(2x) + sin(2x)
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