How do you solve $x'(t)+x3 =0$ ?

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Answer 1 · 2015-08-13T02:24:39

$x = ke^(-3t)$ (for constant $k$ )

$x'(t)+x3=0$ is equivalent to:

$dx/dt +3x=0$ and, in turn:

$dx/x=-3dt$

$int dx/x=-int3dt$

$lnx = -3t +C$

$x = e^(-3t+C)$

$x = e^(-3t)e^C$

$x = ke^(-3t)$ (for constant $k$ )

(Check the answer by differentiating.)

How do you solve x'(t)+x3 =0x'(t)+x3 =0?