Question

(1 point) Find the solution of $$y'' + 2y' = 48 \sin(2t) + 64 \cos(2t)$$ with $$y(0) = 7$$ and $$y'(0) = 7$$. $$y = 133/2-((55/2)e^(-2t))-32\cos(2t)-24\sin(2t)$$

Name: 1 point find the solution of y 2y 48 sin2t 64 cos2t with y0 7 and y0 7 y 1332 552e 2t 32cos2t 24sin2t 23674
Uploaded: 2023-08-18T05:44:25-08:00
Duration: 9 min 4 s
Channel: Adi S
Description: 1 point find the solution of y 2y 48 sin2t 64 cos2t with y0 7 and y0 7 y 1332 552e 2t 32cos2t 24sin2t 23674

          (1 point) Find the solution of
$$y'' + 2y' = 48 \sin(2t) + 64 \cos(2t)$$
with $$y(0) = 7$$ and $$y'(0) = 7$$.
$$y = 133/2-((55/2)e^(-2t))-32\cos(2t)-24\sin(2t)$$

Added by Tanya R.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

08/18/2023

Step 1

The given solution is $$y = \frac{133}{2} - \frac{55}{2}e^{-2t} - 32\cos(2t) - 24\sin(2t)$$ We need to verify if this solution satisfies the differential equation and the initial conditions. First, let's find the first and second derivatives of the given Show more…

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(1 point) Find the solution of $$y'' + 2y' = 48 \sin(2t) + 64 \cos(2t)$$ with $$y(0) = 7$$ and $$y'(0) = 7$$. $$y = 133/2-((55/2)e^(-2t))-32\cos(2t)-24\sin(2t)$$