- #1
karush
Gold Member
MHB
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\tiny{b.2.1.23}
1000
Solve the initial value problem and find the critical value $\displaystyle a_0$ exactly.
$\displaystyle 3y'-2y=e^{-\pi/2}\quad y(0)=a$
divide by 3
$\displaystyle y'-\frac{2}{3}y=\frac{e^{-\pi/2}}{3}$
with integrating factor $\displaystyle e^{-2t/3}$ multiply through
$\displaystyle e^{-2t/3}y'-\frac{2}{3} e^{-2t/3}y=\frac{e^{-\pi/2}}{3}e^{-2t/3}$
ok just seeing if this is going in the right direction
1000
Solve the initial value problem and find the critical value $\displaystyle a_0$ exactly.
$\displaystyle 3y'-2y=e^{-\pi/2}\quad y(0)=a$
divide by 3
$\displaystyle y'-\frac{2}{3}y=\frac{e^{-\pi/2}}{3}$
with integrating factor $\displaystyle e^{-2t/3}$ multiply through
$\displaystyle e^{-2t/3}y'-\frac{2}{3} e^{-2t/3}y=\frac{e^{-\pi/2}}{3}e^{-2t/3}$
ok just seeing if this is going in the right direction
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