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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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Are the functions defined (esp $g$) at $t=0$? Also, I get $W=t$, which is only 0 when $t=0$. –  Daryl Oct 3 '12 at 22:57
    
They certainly are linearly independent. This can be done in several easy ways. –  André Nicolas Oct 3 '12 at 23:02
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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? –  user22805 Oct 3 '12 at 23:08
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2 Answers

up vote 1 down vote accepted

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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Does this allow me to write a general solution for $t^2y'' - ty' + y = 0$? –  Bob John Oct 3 '12 at 23:34
    
@BobJohn:Yes, because you have two linearly independent solutions. You can write the general solution as $y_G(t)=c_1y(t)+c_2g(t)$. –  Mhenni Benghorbal Oct 3 '12 at 23:48
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