MHB Solve IVP 2000 #23: Y(0)=A Solution

[karush]

2000
given #23


so far


I could not get to the W|A solution before applying the y(0)=ahere is the book answer for the rest

 

[skeeter]

$e^{-2t/3} y' - \dfrac{2e^{-2t/3}}{3} y = \dfrac{1}{3}e^{-(3\pi+4)t/6}$

$\left(e^{-2t/3} y \right)' = \dfrac{1}{3}e^{-(3\pi+4)t/6}$

$e^{-2t/3} y = -\dfrac{2}{3\pi+4}e^{-(3\pi+4)t/6} + C$

$y = Ce^{2t/3} -\dfrac{2}{3\pi+4}e^{-\pi t/2}$

 

[karush]

mahalo
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