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#### karush

##### Well-known member

- Jan 31, 2012

- 2,886

\tiny{b.2.1.23}

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

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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