Question

Find the general solution to: y'' + 4y = 8 cos(2t) r^2 + 4 = 0, r = ± 2i y1 = cos(2t), y2(t) = sin(2t) using variation of parameters W(y1, y2) = | cos(2t) sin(2t) ; -2sin(2t) 2cos(2t) | v1' = 1/2 | 0 sin(2t) ; 8cos(2t) 2cos(2t) | = -4cos(2t)sin(2t) = -2sin(4t) v1(t) = 2 cos(4t)/4 = 1/2 cos(4t) v2' = 1/2 | cos(2t) 0 ; -2sin(2t) 8cos(2t) | = 4cos^2(2t) = 2 + 2cos(4t) v2(t) = 2t + sin(4t)/2

          Find the general solution to:
y'' + 4y = 8 cos(2t)
r^2 + 4 = 0, r = &plusmn; 2i
y1 = cos(2t), y2(t) = sin(2t)
using variation of parameters
W(y1, y2) = | cos(2t) sin(2t) ; -2sin(2t) 2cos(2t) |
v1' = 1/2 | 0 sin(2t) ; 8cos(2t) 2cos(2t) | = -4cos(2t)sin(2t) = -2sin(4t)
v1(t) = 2 cos(4t)/4 = 1/2 cos(4t)
v2' = 1/2 | cos(2t) 0 ; -2sin(2t) 8cos(2t) | = 4cos^2(2t) = 2 + 2cos(4t)
v2(t) = 2t + sin(4t)/2

Find the general solution to:
y” + 4y = 8 cos(2t)
r^2 + 4 = 0, r = plusmn; 2i
y1 = cos(2t), y2(t) = sin(2t)
using variation of parameters
W(y1, y2) = | cos(2t) sin(2t) ; -2sin(2t) 2cos(2t) |
v1' = 1/2 | 0 sin(2t) ; 8cos(2t) 2cos(2t) | = -4cos(2t)sin(2t) = -2sin(4t)
v1(t) = 2 cos(4t)/4 = 1/2 cos(4t)
v2' = 1/2 | cos(2t) 0 ; -2sin(2t) 8cos(2t) | = 4cos^2(2t) = 2 + 2cos(4t)
v2(t) = 2t + sin(4t)/2

Added by Jessica M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

05/15/2023

Step 1

First, we have the given differential equation: $$y'' + 4y = 8\cos(2t)$$ We need to find the complementary function and the particular integral to get the general solution. Show more…

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Find the general solution to: y'' + 4y = 8 cos(2t) r^2 + 4 = 0, r = ± 2i y1 = cos(2t), y2(t) = sin(2t) using variation of parameters W(y1, y2) = | cos(2t) sin(2t) ; -2sin(2t) 2cos(2t) | v1' = 1/2 | 0 sin(2t) ; 8cos(2t) 2cos(2t) | = -4cos(2t)sin(2t) = -2sin(4t) v1(t) = 2 cos(4t)/4 = 1/2 cos(4t) v2' = 1/2 | cos(2t) 0 ; -2sin(2t) 8cos(2t) | = 4cos^2(2t) = 2 + 2cos(4t) v2(t) = 2t + sin(4t)/2