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I need to show that $y(t) = t$ and $g(t) = t \ln(t)$ are linearly independent. I thought I could use the Wronskian as follows:

$y'(t) = 1$

$g'(t) = 1 + \ln(t)$

So $W(y, g) = (t)(1 + \ln(t)) - t \ln(t)$, so $W(y, g)(0) = 0$, which means they're not linearly independent. Am I doing something wrong, or is the problem statement written incorrectly?

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    Try it this way. If there were numbers $a$ and $b$ such that $ay(t)+bg(t)$ were everywhere zero, and $a$ and $b$ were not both zero (which is the definition of linear dependence), then what could you say about $\ln t$ as a function? Is this a contradiction?2012-10-03

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The point is that

  • two functions whose Wronskian is nonzero somewhere in an interval where the functions are differentiable are linearly independent on that interval.
  • two functions that are analytic on an open interval $(a,b)$ and whose Wronskian is $0$ everywhere on that interval are linearly dependent on that interval.

It is quite possible to have two analytic functions whose Wronskian is $0$ at some points of the interval but not at others, and then the functions are linearly independent. Consider for example $t$ and $t^2$, whose Wronskian is $0$ at $t=0$.

In your example, the Wronskian of $t$ and $t \ln t$ is in fact nonzero everywhere on the interval $(0,\infty)$ where they are both analytic.

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A related problem. Just apply the idea from linear algebra, two vectors $u_1$ and $u_2$ are linearly independent, if $c_1 u_1 + c_2 u_2 = 0 \Rightarrow c_1=c_2=0\,. $

Apply this to your functions

$ c_1 y(t) + c_2 g(t) = 0 $

Now we need another equation in $c_1$ and $c_2$. We can get it by differentiating the equation with respect to $t$

$ c_1 y'(t) + c_2 g'(t) = 0 \,. $

The two homogeneous equations will have a solution $c_1=c_2=0$ if the determinant $D\neq 0$. That is

$ y(t)g'(t)-g(t)y'(t)=0 \,. $

Applying this to your problem,

$ t(\ln(t) + 1 ) - t\ln(t) = t \neq 0 \,. $

So, the two functions are linearly independent.

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    @BobJohn:Yes, because you have two linearly independent solutions. You can write the general solution as $y_G(t)=c_1y(t)+c_2g(t)$.2012-10-03