Question 1160294
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x'(t)\ =\ -x(t);\ x(0)\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'(t)\ =\ x(t)\ -\ y(t);\ y(0)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z'(t)\ =\ y(t)\ -\ z(t);\ z(0)\ =\ 0]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d(x(t))}{dt}\ =\ -x(t)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d(x(t))}{dt}\ +\ x(t)\ =\ 0]


Let *[tex \Large \mu(t)\ =\ e^{\small{\int\,1\,dt}}\ =\ e^t] and multiply by *[tex \Large \mu(t)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^t\frac{d(x(t))}{dt}\ +\ e^tx(t)\ =\ 0]


Substitute *[tex \Large e^t\ =\ \frac{d}{dt}e^t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^t\frac{d(x(t))}{dt}\ +\ \frac{d}{dt}e^tx(t)\ =\ 0]


Reverse the product rule: *[tex \Large f\frac{dg}{dt}\ +\ g\frac{df}{dt}\ =\ \frac{d}{dt}(fg)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dt}\(e^tx(t)\)\ =\ 0]


Integrate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^tx(t)\ =\ C]


where *[tex \Large C] is an arbitrary constant


Divide both sides by *[tex \Large \mu(t)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(t)\ =\ Ce^{-t}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(0)\ =\ 2\ \Right\ C\ =\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x(t)\ =\ 2e^{-t}]


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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'(t)\ =\ x(t)\ -\ y(t)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y'(t)\ +\ y(t)\ =\ 2e^{-t}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d(y(t))}{dt}\ =\ 2e^{-t}\ -\ y(t)]


Let *[tex \Large \mu(t)\ =\ e^{\small{\int\,1\,dt}}\ =\ e^t] and multiply by *[tex \Large \mu(t)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^t\frac{d(y(t))}{dt}\ +\ e^ty(t)\ =\ 2]



Substitute *[tex \Large e^t\ =\ \frac{d}{dt}e^t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^t\frac{d(y(t))}{dt}\ +\ \frac{d}{dt}e^ty(t)\ =\ 2]


Reverse the product rule: *[tex \Large f\frac{dg}{dt}\ +\ g\frac{df}{dt}\ =\ \frac{d}{dt}(fg)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{dt}\(e^ty(t)\)\ =\ 2]


Integrate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ e^ty(t)\ =\ 2t\ +\ C]


where *[tex \Large C] is an arbitrary constant


Divide both sides by *[tex \Large \mu(t)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(t)\ =\ e^{-t}(2t\ +\ C)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(0)\ =\ 1\ \Right\ C\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(t)\ =\ e^{-t}(2t\ +\ 1)]


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Same process for *[tex \Large z(t)] which yields 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z(t)\ =\ e^{-t}t(t\ +\ 1)]

								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \