# Consider the following linear second-order homogeneous differential equation

Consider the following linear second-order homogeneous differential equation with constant coefficients and two initial conditions
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Korbin Ochoa
Solution:
$\frac{{d}^{2}y\left(t\right)}{{dt}^{2}}-\frac{1}{3}\cdot \frac{dy\left(t\right)}{dt}-\frac{2}{9}\cdot y\left(t\right)=0$

Auxiliary equation:
${m}^{2}-\frac{1}{3}m-\frac{2}{9}=0$
$9{m}^{2}-3m-2=0$
$\left(m-\frac{2}{3}\right)\left(m+\frac{1}{3}\right)=0$ (factor form)

General solution is
$y\left(t\right)={c}_{1}{e}^{\frac{2}{3}\cdot t}+{c}_{2}{e}^{-\frac{1}{3}\cdot t}$ (1)
Apply intral conditions
$y\left(0\right)=0$
$⇒0={c}_{1}{e}^{0}+{c}_{2}{e}^{0}⇒{c}_{1}+{c}_{2}=0$ (2)
and from (1)
$\frac{dy}{dt}\left(t\right)=\frac{2}{3}{c}_{1}{e}^{\frac{2}{3}\cdot t}-\frac{1}{3}{c}_{2}{e}^{-\frac{1}{3}\cdot t}$
Apply $\frac{dy\left(0\right)}{dt}=-3$$⇒-3=\frac{2}{3}{c}_{1}{e}^{0}-\frac{1}{3}{c}_{2}{e}^{0}⇒-9=2{c}_{1}-{c}_{2}$ (3)
and equation (2) and (3)
$2{c}_{1}-{c}_{2}+{c}_{1}+{c}_{2}=-9$
$3{c}_{1}=-9⇒{c}_{1}=-3$
By equation (2) ${c}_{2}=3$
Plug value of ${c}_{1}=-3$ and ${c}_{2}=3$ in equation (1) we get
$y\left(t\right)=-3{e}^{\frac{2}{3}\cdot t}+3{e}^{-\frac{1}{3}\cdot t}$

Jeffrey Jordon

Answer is given below (on video)