Linear dependence/independence of functions $f(x) = x^2+1$, $g(x) = 1+x^3$ and $h(x) = \ln(1 + x)$. Questions :
Prove that the functions $f(x) = x^2+1$, $g(x) = 1+x^3$ and $h(x) = \ln(1 + x)$ are linearly independent on the interval $(0, 1)$. Can you use the Wronskian to do this?
I try this:
$$c_1 x f(x)+ c_2 x g(x) + c_3 x h(x) = 0.$$
but I don't know how to solve for the values of $c_1$, $c_2$, and $c_3$.
Thanks.
 A: Why $c_1 x f(x)+ c_2 x g(x) + c_3 x h(x) = 0$ ?
You have to show that from
$(*)$ $c_1 f(x)+ c_2  g(x) + c_3  h(x) = 0$ for all $x \in (0,1)$ 
it follows that $c_1=c_2=c_3=0$.
From $(*)$ it follows by continuity:
$c_1 f(x)+ c_2  g(x) + c_3  h(x) = 0$ for all $x \in [0,1]$ .
Now let $x=0$, $x=1$ and $x=1/2$, then you get a very easy linear system for the coefficients $c_i$.
It is your turn to show:  $c_1=c_2=c_3=0$.
A: The Wronskian is \begin{align}W(f,g,h)(x)&=\begin{vmatrix}f(x)& g(x) & h(x)\\f'(x) & g'(x) & h'(x) \\f''(x) & g''(x)& h''(x)\end{vmatrix}=\begin{vmatrix}1+x^2& 1+x^3 & \ln{(1+x)}\\2x & 3x^2 & \frac1{1+x} \\2 & 6x & -\frac{1}{(1+x)^2}\end{vmatrix}\\[0.3cm]&=\left\{\text{terms without} \ln(x+1)\right\}+6x^2\ln(x+1)\end{align} which means that the Wronskian does not vanish, hence the functions are linearly independent.
A: $$f(x)=x^2+1$$
$$h(x)=1=x^3$$
$$h(x)=\ln(1+x)$$
$$c_1xf(x)+c_2xg(x)+c_3xh(x)=0$$
Dividing through $x$,
We have got
$$c_1f(x)+c_2g(x)+c_3h(x)=0$$
There is a theorem states that if the constants 
$$c_1=c_2=c_3=0$$
it follows that the vector is linearly independent.
$$c_1(x^2+1)+c_2(1+x^3)+c_3(\ln(1+x))=0$$
$$c_1x^2+c_2x^3+c_1+c_2+c_3\ln(1+x)=0$$
It is obvious that the trivial solution is 
$$c_1=c_2=c_3=0$$
since 
the coefficients of $x^2$ which is $c_1=0$. The argument for the rest also follows.
