Question

Having second guesses on this one...
Solving the Linear system of Differential Equations

Answer 1

\[\large x' = \left[\begin{matrix}14 & -16 \\ 12 & -14\end{matrix}\right]x\]
Let \(\large a = 14, b = -16, c = 12, d = -14\)

\[\large \left\{ x_1' = ax_1 + bx_2\} \\ \{x_2' = cx_1 + dx_2 \right\}\]
Derivative of \(x_1'\)

\[\large x''_1 = ax_1' + bx'_2\]
plugged in \(x'_2 \) from first equation
\[\large x''_1 = ax_1' + b(cx_1 + dx_2)\]
\[\large x''_1 = ax_1' + bcx_1 + bdx_2\]
Now plugged in \( bx_2 \) from first equation
\[\large x''_1 = ax_1' + bcx_1 + d(x'_1 - ax_1)\]
\[\large x''_1 = ax_1' + bcx_1 + dx'_1 - dax_1\]
\[\large x''_1 = (a + d)x_1' + (bc - da)x_1\]

Now plugging in numbers

\[\large x''_1 = \cancel{(14 -14)x_1'} + ((-16 \times 12) - (-14 \times 14)x_1\]

\[\large x''_1 = \cancel{(14 -14)x_1'} + ((-16 \times 12) - (-14 \times 14)x_1\]

\[\large x''_1 = (192) - (-196)x_1\]
\[\large x''_1 = (-192 + 196)x_1\]
\[\large x''_1 = 4x_1\]
set = to 0
\[\large x''_1 - 4x_1 = 0\]
\[\large = \pm 2 \]

so \(\large x_1 = C_1e^{2t} + C_2e^{-2t} \)

Answer 2

Now take the derivative of that
\[\large x_1' = 2C_1e^{2t} - 2C_2e^{-2t}\]
plug in \(x'_1\)
\[\large 14x_1 - 16x_2 = 2C_1e^{2t} - 2C_2e^{-2t}\]
Plug in x_1 we just found
\[\large 14(C_1e^{2t} + C_2e^{-2t}) - 16x_2 = 2C_1e^{2t} - 2C_2e^{-2t}\]
\[\large 14C_1e^{2t} + 14C_2e^{-2t} - 16x_2 = 2C_1e^{2t} - 2C_2e^{-2t}\]
solve for \( x_2 \)
\[\large 14C_1e^{2t} + 14C_2e^{-2t} - 2C_1e^{2t} + 2C_2e^{-2t} = 16x_2\]
\[\large 12C_1e^{2t} + 16C_2e^{-2t} = 16x_2\]
\[\large x_2 = \frac{3}{4}C_1e^{2t} + C_2e^{-2t}\]

Answer 3

I let \(\large c_1 = 4C_1 \) so

\[\large x_1 = 4C_1e^{2t} + C_2e^{-2t}\]
\[\large x_2 = 3C_1e^{2t} + C_2e^{-2t}\]

Answer 4

Typed it out beforehand lol...

Does that look about right?

Answer 5

yeah everything looks good to me, don't see any mistakes

here is what wolfram gives, which is equivalent since constants are arbitrary
http://www.wolframalpha.com/input/?i=x%27+%3D+14x+-16y+%2C+y%27+%3D+12x+-14y

Answer 6

I always forget about wolfram! Thanks @dumbcow

Answer 7

haha :)