# Math Help - behavior of solutions

1. ## behavior of solutions

If a>0, b>0, c=0, show that all solutions of ay''+by'+cy=0 approach a constant that depends on the initial conditions as t approaches to infinity. Determine this constant for the initial conditons $y(0)=y_0, y'(0)=y_0'$
-------------------------------------------------------
my approach:
$y=c_1e^{r_1 t}+c_2e^{r_2 t}$
$b^2-4ac=b^2$
$r_1=0, r_2= \frac{-b-b}{2a}$
$y=c_1e^0+c_2e^{\frac{-bt}{a}}$
$y(0)=c_1=y_0$
$y'(t)=c_2(-{\frac{b}{a})e^{\frac{-bt}{a}}}$
$t=0, c_2=y_0'(-{\frac{a}{b}})$
$y=c_1e^0+c_2e^{\frac{-bt}{a}}$
$=c_1+c_2e^{\frac{-bt}{a}}$
$=y_0+{y_0'}(-{\frac{a}{b}})e^{\frac{-bt}{a}}$
which approaches to $y_0$ as t approaches to infinity.

(according to the solution manual, the constant should be $y_0+{\frac{a}{b}}y_0'$ )

2. Hello elmo
Originally Posted by elmo
If a>0, b>0, c=0, show that all solutions of ay''+by'+cy=0 approach a constant that depends on the initial conditions as t approaches to infinity. Determine this constant for the initial conditons $y(0)=y_0, y'(0)=y_0'$
-------------------------------------------------------
my approach:
$y=c_1e^{r_1 t}+c_2e^{r_2 t}$
$b^2-4ac=b^2$
$r_1=0, r_2= \frac{-b-b}{2a}$
$y=c_1e^0+c_2e^{\frac{-bt}{a}}$
$\color{red}y(0)=c_1=y_0$
$y'(t)=c_2(-{\frac{b}{a})e^{\frac{-bt}{a}}}$
$t=0, c_2=y_0'(-{\frac{a}{b}})$
$y=c_1e^0+c_2e^{\frac{-bt}{a}}$
$=c_1+c_2e^{\frac{-bt}{a}}$
$=y_0+{y_0'}(-{\frac{a}{b}})e^{\frac{-bt}{a}}$
which approaches to $y_0$ as t approaches to infinity.

(according to the solution manual, the constant should be $y_0+{\frac{a}{b}}y_0'$ )

See the line I've highlighted in red. Can you see your mistake? $e^0=1$, not $0$.