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For the system

$$x' = \left[ \begin{array}{cccc} 2&6\\3&-1 \end{array} \right]x$$

with solutions

$$x_1 = \left[ \begin{array}{cccc} 2e^{5t}\\e^{5t} \end{array} \right]\qquad \text{and}\quad x_2 = \left[ \begin{array}{cccc} e^{-4t}\\-e^{-4t} \end{array} \right]$$

A. Use the Wronskian to show that the solutions are linearly independent.

B. Write the general solution to the problem.

For A, would I just compute the determinant of the matrix $$x = \left[ \begin{array}{cccc} 2e^{5t}&e^{-4t}\\e^{5t}&-e^{-4t} \end{array} \right]$$ and show that it never equals zero, thus proving that the two solutions are linearly independent?

For B, would the general solution just be $$x(t) = c_1\left[ \begin{array}{cccc} 2e^{5t}\\e^{5t} \end{array} \right] + c_2\left[ \begin{array}{cccc} e^{-4t}\\-e^{-4t} \end{array} \right]~?$$

Thanks!

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For linear independence, it's sufficient for the determinant to not vanish identically, i.e. to not vanish for all values of $t$. For example, the two functions $t \to (1,t)$ and $t \to (1,t^2)$ are linearly independent even though $\left|\begin{smallmatrix} 1 & 1 \\ t & t^2 \end{smallmatrix}\right|$ vanishes for $t=0$.

In your case, however, the distinction is moot. The RHS (i.e. $\left(\begin{smallmatrix} 2 & 6 \\ 3 & -1 \end{smallmatrix}\right)x$) of your ODE satisfies a global lipschitz condition (as do the right-hand sides of all homogenous linear ODEs with constant coefficients), which makes the solution globally unique. This means that for every possible choice of $t$ and every initial condition $x(t) = (a_0,a_1)$, there is exactly one solution of the ODE. If there were $c_1,c_2$ with $c_1 x_1(t_0) + c_2 x_2(t_0) = 0$ for some $t_0$ but with $c_1 x_1(t_1) + c_2 x_2(t_1) \neq 0$ for some $t_1$, then both $$ c_1x_1(t) \text{ and } -c_2x_2(t) $$ would be solutions of the initial value problem $$ x(t_0) = c_1x_1(t_0) \quad \text{(or equivalently } x(t_0) = -c_2x_2(t_0) \text{ )} $$ yet the two solutions would not be identical since you'd have $$ c_1x_1(t_1) \neq -c_2x_2(t_1) $$

Your check for linear independence is thus OK. As is your statement about the general solution.

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  • $\begingroup$ Anything for B? $\endgroup$ – user1038665 Nov 7 '12 at 10:50
  • $\begingroup$ @user1038665 (B) is fine too. $\endgroup$ – fgp Nov 7 '12 at 11:36

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