# [SOLVED]b.2.1.23 Solve the initial value problem and find the critical value a0 exactly.

#### karush

##### Well-known member
\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

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#### Country Boy

##### Well-known member
MHB Math Helper
It's easy to check. If $e^{-2t/3}$ is an integrating factor then $\frac{d}{dt}(e^{-2t/3}y)= e^{-2t/3}y'- \frac{2}{3}e^{-2t/3}y= e^{-2t/3}\frac{e^{-\pi/2}}{3}$ so, yes, that is the correct integrating factor.

To complete, integrate both sides of $\frac{d}{dt}\left(e^{-2t/3}y\right)= e^{-(4t- 3\pi)/6}$ which is, of course, the same as $d\left(e^{-2t/3}y\right) =\left(e^{-(4t- 3\pi)/6}\right)dt$ including the constant of integration, using the condition that y(0)= 1 to determine the value of that constant.

Since this is a "linear equation with constant coefficients" it can also be done, perhaps more easily, by solving the "characteristic equation", $3r- 2= 0$ so $r= \frac{2}{3}$ meaning that the general solution to the "associated homogeneous equation", 3y'- 2y= 0, is $y_h(t)= Ce^{2t/3}$ where C is an undetermined constant. We then look for a single function to satisfy the entire equation $3y'- 2y= e^{-\pi/2}$. Since the right side is a contant, we try y= A so that y'= 0 and the equation becomes $-2A- e^{-\pi/2}$ so that $A= -\frac{1}{2}e^{-pi/2}$ and the general solution to the entire equation is $y(t)= Ce^{2t/3}- \frac{1}{2}e^{-\pi/2}$. Use $y(0)= C- \frac{1}{2}e^{-\pi/2}= 1$ to determine the value of C.

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