Question

d^2y/dx^2 - 4(dy/dx) + 4y = 0; y = c1e^(2x) + c2xe^(2x) When y = c1e^(2x) + c2xe^(2x), (dy/dx) = d^2y/dx^2 = 2c1e^(2x) + 2c2e^(2x). Thus, in terms of x, d^2y/dx^2 - 4(dy/dx) + 4y = 2c1e^(2x) + 2c2e^(2x) - 4(2c1e^(2x) + 2c2e^(2x)) + 4(c1e^(2x) + c2xe^(2x)) = 0.

Name: d2y dx2 4 dy dx 4y 0 y c1e2x c2xe2x when y c1e2x c2xe2x dy dx d2y dx2 thus in terms of x d2y dx2 4 dy dx 4y 4c1e2x c2xe2x 77034
Uploaded: 2023-07-17T16:22:59-08:00
Duration: 4 min 42 s
Channel: Adi S
Description: d2y dx2 4 dy dx 4y 0 y c1e2x c2xe2x when y c1e2x c2xe2x dy dx d2y dx2 thus in terms of x d2y dx2 4 dy dx 4y 4c1e2x c2xe2x 77034

          d^2y/dx^2 - 4(dy/dx) + 4y = 0; y = c1e^(2x) + c2xe^(2x)
When y = c1e^(2x) + c2xe^(2x),
(dy/dx) = d^2y/dx^2 = 2c1e^(2x) + 2c2e^(2x).
Thus, in terms of x,
d^2y/dx^2 - 4(dy/dx) + 4y = 2c1e^(2x) + 2c2e^(2x) - 4(2c1e^(2x) + 2c2e^(2x)) + 4(c1e^(2x) + c2xe^(2x)) = 0.

Added by Brian W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/17/2023

Step 1

(dy/dx) = 2c1e^(2x) + c2e^(2x) + 2c2xe^(2x) Show more…

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d^2y/dx^2 - 4(dy/dx) + 4y = 0; y = c1e^(2x) + c2xe^(2x) When y = c1e^(2x) + c2xe^(2x), (dy/dx) = d^2y/dx^2 = 2c1e^(2x) + 2c2e^(2x). Thus, in terms of x, d^2y/dx^2 - 4(dy/dx) + 4y = 2c1e^(2x) + 2c2e^(2x) - 4(2c1e^(2x) + 2c2e^(2x)) + 4(c1e^(2x) + c2xe^(2x)) = 0.