Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For the system

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

with solutions

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

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 $x(t) = c_1\left[ \begin{array}{cccc} 2e^{5t}\\e^{5t} \end{array} \right] + c_2\left[ \begin{array}{cccc} e^{-4t}\\-e^{-4t} \end{array} \right]$?


share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
Anything for B? – user1038665 Nov 7 '12 at 10:50
@user1038665 (B) is fine too. – fgp Nov 7 '12 at 11:36
Thanks! Appreciate it :) – user1038665 Nov 7 '12 at 20:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.