So I got this classic implicit differentiation problem y=x^x I've tried to tackle it by transforming x^x to the equivalent form of e^((lnx)x): y=xx y=(e^(lnx))^x y=e^((lnx)x) and then proceeded to differentiate it in the following way: y′=e:((lnx)x) x lnx x 1/x

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The solution of a differential equation y′′+3y′+2y=0 is of the form
A) $c_1{e}^x+c_2{e}^{2x}$
B) $c_1{e}^{-<!--\; -\; -->x}+c_2{e}^{3x}$
C) $c_1{e}^{-<!--\; -\; -->x}+c_2{e}^{-<!--\; -\; -->2x}$
D) $c_1{e}^{-<!--\; -\; -->2x}+c_2{2}^{-<!--\; -\; -->x}$

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A. 4.2
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C. 4.7
D. 5.1

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