Hw$08-2\mathrm{.}3-$SODE$-$CCE: Problem $8$

Problem Value: $5$ point$($s$).$ Problem Score: $0\%.$ Attempts Remaining: $3$ attempts.

Help Entering Answers See Examples $2\mathrm{.}3\mathrm{.}1-2\mathrm{.}3\mathrm{.}5,$ in Section $2\mathrm{.}3,$ in the MTH $235$ Textbook.

$(5$ points$)$

You have only $4$ attempts in this problem

Homework: Homogeneous Constant Coefficients

Find $y(t)$ solution of the initial value problem

$y^{\text{'}\text{'}}-y^{\text{'}}-6y=0,y(0)=y_0,y^{\text{'}}(0)=y_1$

where $y_0$ and $y_1$ are arbitrary constants.

A$.y(t)=y_1{e}^{3t}{y}_0{e}^{-2t}$

B$.y(t)=\frac{1}{2}(y_0{y}_1)e^{3t}\frac{1}{2}(y_0-y_1)e^{-2t}$

C$.y(t)=y_1{e}^{-3t}{y}_0{e}^{2t}$

D$.y(t)=\frac{1}{5}(2y_0{y}_1)e^{-3t}\frac{1}{5}(3y_0-y_1)e^{2t}$

E$.y(t)=y_0{e}^{3t}{y}_1{e}^{-2t}$

F$.y(t)=\frac{1}{5}(2y_0{y}_1)e^{3t}\frac{1}{5}(3y_0-y_1)e^{-2t}$

G$.y(t)=\frac{1}{2}(y_0{y}_1)e^{-3t}\frac{1}{2}(y_0-y_1)e^{2t}$

H$.y(t)=\frac{1}{5}(2y_0-y_1)e^{-3t}\frac{1}{5}(3y_0{y}_1)e^{2t}$

$y(t)=\frac{1}{5}(2y_0-y_1)e^{3t}\frac{1}{5}(3y_0{y}_1)e^{-2t}$

J$.y(t)=y_0{e}^{-3t}{y}_1{e}^{2t}$

K$.$ None of the above.