### Author Topic: TUT0401 QUIZ5  (Read 2135 times)

#### Xi Zheng

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##### TUT0401 QUIZ5
« on: November 17, 2019, 04:46:30 PM »
Find a particular solution of the given inhomogeneous Euler's euqation:
(1 - t)$y\prime\prime$ + t$y\prime$ - y = $2(t - 1)^2$$e^{-t}, 0 < t < 1, y_1(t) = e^t. Solution: y_1 = e^{t}, y_1\prime = e^{t} Rewrite: y\prime\prime + \frac{t}{1-t}$$y\prime$ - $\frac{y}{1 - t}$ = $\frac{2(t-1)^2e^{-t}}{1-t}$

p(t) = $\frac{t}{1-t}$

g(t) = 2$\frac{(t-1)^2e^{-t}}{1-t}$

$y_1$$v\prime\prime + [2y_1\prime + p(t)y_1]v\prime = g(t) e^t$$v\prime\prime$ + [2$e^t$+ $\frac{t}{1-t}$$e^t]v\prime = 2\frac{(t-1)^2e^{-t}}{1-t} v\prime\prime + [\frac{2-t}{1-t}]v\prime = 2\frac{(t-1)^2e^{-t}}{1-t} Let w = v\prime: w\prime + [\frac{2-t}{1-t}]w = 2\frac{(t-1)^2e^{-t}}{1-t} \mu(t) = e^{\int \frac{2-t}{1-t}} = \frac{e^t}{1-t} \frac{e^t}{1-t}$$w\prime$ + $\frac{e^t}{1-t}$[$\frac{2-t}{1-t}$]w = 2$\frac{(t-1)^2e^{-t}}{1-t}$$\frac{e^t}{1-t} \mu(t)w = \int \frac{2(t-1)^2e^{-t}}{(1-t)^2} = \int 2e^{-t} = -2e^{-t} w = -2e^{-t}$$\frac{1-t}{e^t}$ = -2(1-t)$e^{-2t}$

w = $v\prime$ = -2(1-t)$e^{-2t}$

v = $\int -2(1-t)e^{-2t}$

By using substitution, we get:
v = (1-t)$e^{-2t}$- $\frac{1}{2}$$e^{-2t} Y(t) = vy_1 = (1-t) e^{-t}- \frac{1}{2}$$e^{-2t}$

« Last Edit: November 17, 2019, 05:01:50 PM by zhengxi »