Question

Find the solution $y(t)$ of the given IVP: $6y'' + y' - 2y = 0$, $y(0) = 1$, $y'(0) = 4$ $\bigcirc y(t) = 2e^{\frac{1}{2}t} - 4e^{-\frac{1}{3}t}$ $\bigcirc y(t) = 3e^{\frac{1}{3}t} - e^{-\frac{1}{2}t}$ $\bigcirc y(t) = 4e^{\frac{1}{3}t} - 3e^{-\frac{1}{2}t}$ $\bigcirc y(t) = 5e^{-\frac{1}{3}t} - 2e^{\frac{1}{2}t}$

          Find the solution $y(t)$ of the given IVP:
$6y'' + y' - 2y = 0$, $y(0) = 1$, $y'(0) = 4$
$\bigcirc y(t) = 2e^{\frac{1}{2}t} - 4e^{-\frac{1}{3}t}$
$\bigcirc y(t) = 3e^{\frac{1}{3}t} - e^{-\frac{1}{2}t}$
$\bigcirc y(t) = 4e^{\frac{1}{3}t} - 3e^{-\frac{1}{2}t}$
$\bigcirc y(t) = 5e^{-\frac{1}{3}t} - 2e^{\frac{1}{2}t}$

Find the solution y(t) of the given IVP:
6y” + y' - 2y = 0, y(0) = 1, y'(0) = 4
◯ y(t) = 2e^(1)/(2)t - 4e^-(1)/(3)t
◯ y(t) = 3e^(1)/(3)t - e^-(1)/(2)t
◯ y(t) = 4e^(1)/(3)t - 3e^-(1)/(2)t
◯ y(t) = 5e^-(1)/(3)t - 2e^(1)/(2)t

Added by Joanne F.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

09/02/2023

Step 1

Let's rewrite this in standard form by letting y = y(t), y' = dy/dt, and y'' = d^2y/dt^2. The equation becomes: 6y'' + y' - 2y = 0 6(d^2y/dt^2) + dy/dt - 2y = 0 Show more…

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Find the solution y(t) of the given IVP: 6y'' + y' - 2y = 0, y(0) = 1, y'(0) = 4 y(t) = 2e^(2/3)t - 4e^(1/2)t y(t) = 3e^(2/3)t - e^(4/3)t y(t) = 4e^(1/2)t - 3e^(-2/3)t y(t) = 5e^(-1/2)t - 2e^(2/3)t Find the solution y(t) of the given IVP: 6y + y - 2y = 0, y0 = 1, 0 = 4 Oyt = 2ei - 4e Oyt = 3ei - c Oy = 4e - 3e -1g0