Mathematics, 15.01.2020 17:31, jaylanmahone223
4. show that both x1(t) = e−tcos tandx2(t) = e−tsin t are solutions to the second-order differential equation x′′+ 2x′+ 2x = 0. show that x(t) = ae−tsin t + be−tcos t is a solution for any values of the constants a and b
Answers: 3
![[tex]x_2(t) = e^{-t}\sin t\\x_2'(t) = -e^{-t}\sin t+e^{-t}\cos t\\x_2''(t) = e^{-t}\sin t-e^{-t}\cos t-e^{-t}\cos t-e^{-t}\sin t=-2e^{-t}\cos t\\\\x''+2x'+2x =-2e^{-t}\cos t-2e^{-t}\sin t+2e^{-t}\cos t+2e^{-t}\sin t=0](/tpl/images/0456/2176/51b6a.png)
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4. show that both x1(t) = e−tcos tandx2(t) = e−tsin t are solutions to the second-order differential...
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