Question

use Laplace transforms to solve the system: y(dot) +6y=x(dot), 3x-x(dot)=2y(dot); x(0)=2, y(0)=3

Answer 1

What do dot represent?

Answer 2

the derivatives over time so x(dot)=dx/dt and y(dot)=dy/dt

Answer 3

y'+6y=x'
3x-x'=2y'

sY(s)-3 + 6 Y(s)= s X(s)-2
3X(s)-(S X(s)-2)= 2 sY(s)- 3
x(0)=2
y(0)=3

Answer 4

Well, we got two equation , two unknown

Answer 5

Have to solve for X(s) and Y(s)

Answer 6

i cant seem to get an answer keep on getting stuck.. Could you help me out?

Answer 7

good you are still here

Answer 8

Check my algebra

Y (s) (s + 6) - s X (s) = 1
Y (s) (-2 s) + (3 - s) X (s) = -5

Answer 9

\[\left(
\begin{array}{cc}
s+6 & -s \\
-2 s & 3-s
\end{array}
\right)\left(
\begin{array}{c}
Y(s) \\
X(s)
\end{array}
\right)=\left(
\begin{array}{c}
1 \\
-5
\end{array}
\right)\]

Answer 10

A x =B

x= A^-1 B

Answer 11

made a stupid error there.. wrote -1 instead of -5

Answer 12

When solved
Y(s)=\[\frac{3-s}{18-3 s-3 s^2}-\frac{5 s}{18-3 s-3 s^2}\]

Answer 13

X(s)=\[\frac{2 s}{18-3 s-3 s^2}-\frac{5 (6+s)}{18-3 s-3 s^2}\]

Answer 14

y(t)=\[\frac{1}{5} e^{-3 t} \left(7+3 e^{5 t}\right)\]

x(t)=\[\frac{1}{5} e^{-3 t} \left(-7+12 e^{5 t}\right)\]

Answer 15

Let's check
\[y'[t]+6y[t]=\frac{3}{5} e^{-3 t} \left(7+8 e^{5 t}\right)\]

\[x'[t]=\frac{3}{5} e^{-3 t} \left(7+8 e^{5 t}\right)\]

They Match!!!!, We are good

Answer 16

wow you are simply amazing :)