Show that $y(t) = t$ and $g(t) = t \ln(t)$ are linearly independent 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?
 A: 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. 
A: 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.
