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

karush · Jun 17, 2019

2000
given #23

Printed list of differential equation initial value problems: (b) solve y' - (1/2)y = 2 cos t with y(0) = a and find the critical value a0; (c) describe the solution behavior; 21. y' - (1/2)y = 2 cos t, y(0) = a; 22. 2y' - y = e^(t/2), y(0) = a; 23. 3y' - 2y = e^(-t/2), y(0) = a.

so far

Worked solution to problem 2.1.23: solving 3y' - 2y = e^(-πt/2) with y(0) = a, dividing by 3, finding integrating factor e^(-2t/3), and final Wolfram Alpha form y = c1 e^(2t/3) - 2e^(-πt/2)/(4 + 3π).

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

skeeter · Jun 17, 2019

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

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

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

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

karush · Jun 22, 2019

mahalo
noticed there was 430+ views on this one

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

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