Question

Question 1 Let $y(t)$ is the solution of the initial value problem $y'' + 5^2y = ksin(4t)$ with $y(0) = 0$, $y'(0) = 0$. If $y(\frac{\pi}{2}) = 3$, then $k$ is equal to is equal to: -37.25 -38.75 -33.75 -31.25 -29.75

          Question 1
Let $y(t)$ is the solution of the initial value problem $y'' + 5^2y = ksin(4t)$ with $y(0) = 0$, $y'(0) = 0$. If $y(\frac{\pi}{2}) = 3$, then $k$ is equal to is equal to:
-37.25
-38.75
-33.75
-31.25
-29.75

Question 1
Let y(t) is the solution of the initial value problem y” + 5^2y = ksin(4t) with y(0) = 0, y'(0) = 0. If y((π)/(2)) = 3, then k is equal to is equal to:
-37.25
-38.75
-33.75
-31.25
-29.75

Added by Javier D.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ishana K

04/23/2022

Step 1

Step 1: We are given the initial value problem y"+52y=ksin(4t) with y0=0 and y(0)=0. Show more…

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Question 1 Let y(t) is the solution of the initial value problem y"+52y=ksin(4t) with y0=0, y(o=0. If y=3. then is equal to is equal to: O -37.25 O -38.75 O -33.75 O -31.25 O -29.75