Question

$r^2 - 4r + 3 = 0$ $y_1 = e^t$ $y_2 = e^{3t}$ $W = 2e^{4t}$ $\int \frac{y_1 g}{W} dt = e^{-2t}$ $\int \frac{y_2 g}{W} dt = -2t$ $c_1 e^{3t} + (c_2 - 4)e^t$ $y'' - 4y' + 3y = -4e^t$ $y = c_1 e^{3t} + c_2 e^t$

Name: r2 4r 3 0 y1 et y2 e3t w 2e4t int fracy1 gw dt e 2t int fracy2 gw dt 2t c1 e3t c2 4et y 4y 3y 4et y c1 e3t c2 et 37947
Uploaded: 2022-03-22T06:52:01-08:00
Duration: 4 min 53 s
Channel: Sutara K
Description: r2 4r 3 0 y1 et y2 e3t w 2e4t int fracy1 gw dt e 2t int fracy2 gw dt 2t c1 e3t c2 4et y 4y 3y 4et y c1 e3t c2 et 37947

          $r^2 - 4r + 3 = 0$
$y_1 = e^t$
$y_2 = e^{3t}$
$W = 2e^{4t}$
$\int \frac{y_1 g}{W} dt = e^{-2t}$
$\int \frac{y_2 g}{W} dt = -2t$
$c_1 e^{3t} + (c_2 - 4)e^t$
$y'' - 4y' + 3y = -4e^t$
$y = c_1 e^{3t} + c_2 e^t$

$r2 4r 3 0 y1 et y2 e3t w 2e4t int fracy1 gw dt e 2t int fracy2 gw dt 2t c1 e3t c2 4et y 4y 3y 4et y c1 e3t c2 et 37947$

Added by Christina B.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Sutara K

03/22/2022

Step 1

This factors as $(r-1)(r-3) = 0$, so the roots are $r_1 = 1$ and $r_2 = 3$. The fundamental solutions are $y_1 = e^t$ and $y_2 = e^{3t}$. Show more…

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$r^2 - 4r + 3 = 0$ $y_1 = e^t$ $y_2 = e^{3t}$ $W = 2e^{4t}$ $\int \frac{y_1 g}{W} dt = e^{-2t}$ $\int \frac{y_2 g}{W} dt = -2t$ $c_1 e^{3t} + (c_2 - 4)e^t$ $y'' - 4y' + 3y = -4e^t$ $y = c_1 e^{3t} + c_2 e^t$