I'm trying to solve the following problem from my textbook, but I don't have the answer. I just want to know if I'm on the right path (since this topic was skimmed over in my last lecture).

Given: $x_1'=-7x_1+9x_2$, $x_2'=-9x_1-7x_2$, and initial conditions $x_1(0)=3, x_2(0)=4$, solve for $x_1$ and $x_2$

Solve for $x_2$ in terms of $x_1$: $$9x_2=x_1'+7x_1$$ $$x_2=\frac{1}{9}x_1'+\frac{7}{9}x_1$$

Solve for $x_2'$ using the above equation: $$x_2'=\frac{1}{9}x_1''+\frac{7}{9}x_1'$$

Plug these into the second equation: $$x_2'=-9x_1-7x_2$$ $$\frac{1}{9}x_1''+\frac{7}{9}x_1'=-9x_1-7(\frac{1}{9}x_1'+\frac{7}{9}x)$$

Organize to get the general form of a second-order differential equation: $$\frac{1}{9}x_1''+\frac{14}{9}x_1'+\frac{130}{9}x_1=0$$

I personally hate fractions, so I multiplied by nine at this point: $$x_1''+14x_1'+130x_1=0$$

Turn this into a parabolic equation: $$r^2+14r+130$$

Solve for r, get that it's complex: $r=-7+/-9i$

Now, $$x_1=e^{-7t}(c_1\cos(9t)+c_2\sin(9t))$$ $$x_1'=-7e^{-7t}(c_1\cos(9t)+c_2\sin(9t))+9e^{-7t}(-c_1\sin(9t)+c_2\cos(9t))$$

Therefore, $$x_2=\frac{1}{9}(-7e^{-7t}(c_1\cos(9t)+c_2\sin(9t))+9e^{-7t}(-c_1\sin(9t)+c_2\cos(9t)))+\frac{7}{9}(e^{-7t}(c_1\cos(9t)+c_2\sin(9t)))$$

Solving for $c_1$ using the initial conditions: $$x_1(0)=3=e^{-7(0)}(c_1\cos(9(0))+c_2\sin(9(0)))$$ $$3=c_1$$

Solving for $c_2$: $$x_2(0)=4=\frac{1}{9}(-7e^{-7(0)}(3\cos(9(0))+c_2\sin(9(0)))+9e^{-7(0)}(-3\sin(9(0))+c_2\cos(9(0))))+\frac{7}{9}(e^{-7(0)}(3\cos(9(0))+c_2\sin(9(0))))$$ $$4=\frac{1}{9}(3+9(c_2))$$ $$36=3+9c_2$$ $$\frac{33}{9}=c_2$$

Therefore, $3=c_1$, $\frac{33}{9}=c_2$

Is this correct?


$x_1' = -7x_1 + 9x_2\\ x_2' = -9x_1 - 7x_2$

Option 1:

$A = \pmatrix{-7&9\\-9&-7}\pmatrix{x_1\\x_2}\\ \pmatrix{x_1\\x_2}' = A\pmatrix{x_1\\x_2}\\ \mathbf x = e^{At}\mathbf x_0$

if $A = P^{-1}DP\\e^{At} = P^{-1}e^{Dt}P\\ P\mathbf x = e^{Dt} P\mathbf x_0$

$\lambda = -7 \pm 9i$

$e^{At} = \pmatrix {e^{-7t}\cos 9t & e^{-7t}\sin 9t\\-e^{-7t}\sin 9t&e^{-7t}\cos 9t}$

$x_1 = 3 e^{-7t}\cos 9t + 4 e^{-7t}\sin 9t\\ x_2 = 4 e^{-7t}\cos 9t - 3 e^{-7t}\sin 9t$

Option 2:

$9x_2 = x_1' + 7 x_1\\ 9x_2' = x_1'' + 7 x_1'\\ -81 x_1 - 63 x_2 = x_1'' + 7x_1'\\ -81 x_1 - 7(x_1' + 7x_1) = x_1'' + 7 x_1'\\ x_1'' + 14 x_1' + 130 x_1 = 0$

Now you have a second order differential equation.

$x_1 = Ae^{-7t}\cos 9x + B e^{-7}\sin 9x$

initial conditions:

$x_1 = 3e^{-7t}\cos 9x + B e^{-7}\sin 9x\\ x_1' = -7x_1 - 9\cdot 3 e^{-7t}\sin 9x + 9\cdot B e^{-7t}\cos 9x = -7x_1 + 9x_2\\ x_2 = B e^{-7t}\cos 9x - 3 e^{-7t}\sin 9x $

Initial conditions indicate $B = 4$


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