While getting into Differential Equations I came across an exercise:
Find the solutions of $ty'' +y'=0$ and prove that they are linearly independent.
This is a 2nd Order ordinary differential eqution therefore my approach to find the solutions was:
Set $y'=v(t) \Rightarrow y''=v'(t)=\frac{dv(t)}{dt}$
Then through substituting we get:
$$ \frac{dv(t)}{dt} = -\frac{v(t)}{t} \Rightarrow \frac{\frac{dv(t)}{dt}}{v(t)}=-\frac{v(t)}{tv(t)} $$ $$\frac{\frac{dv(t)}{dt}}{v(t)} = -\frac{1}{t} \Rightarrow \int{\frac{\frac{dv(t)}{dt}}{v(t)}dt}=\int{-\frac{1}{t}dt}$$ which equals to: $$ ln(v(t))=-ln(t)+c_1$$ where c1 is an arbitrary constant. By solving for v(t) we get: $$v(t)=\frac{e^{c_1}}{t} \Rightarrow y'=\frac{e^c-1}{t}$$ So we get that $$ y=\int{\frac{{e^{c_1}}}{{t}}dt}$$ Where because $c_1$ an arbitrary constant is, we can substitute $e^{c_1}$ with $c_1$ And the final solution is $y=c_1 ln(t) +c_2$ So to my understanding all solutions of this DE have this form. In order to prove the linear independence my approach was to use the Wronskian determinant, and through proving that it's never equal to zero the solutions are linearly independent.
When I tried this approach, I chose two arbitrary solutions, namely: $ c_1ln(t) +c_2 $ and $ c_3ln(t) +c_4$. So the Wronskian was: $$|W| =\begin{vmatrix}c_1ln(t)+c_2&c_3ln(t)+c_4\\\frac{c_1}{t}&\frac{c_3}{t}\end{vmatrix}$$ Which equals to:$ |W|= \frac{c_2c_3-c_4c_1}{t}$ Which is equal to zero for $c_2c_3=c_4c_1$ and therefore my thinking is wrong. This is where I am stuck and I don't know what is wrong in my approach. Any insight would be helpful because I think I am confusing myself with something trivial. Since this DE is homogeneous it also has the trivial solution $y=0$ but is it valid to plug this solution in the wronskian?
$$ty'' +y'=0$$ Rewrite the DE and integrate twice: $$(ty')'=0$$ $$ty'=C_1$$ $$y(t)=C_1 \color {red}{ \times \ln (t)}+C_2 \color {red}{\times 1}$$ For the Wronskian you have that: $$W(\ln t,1)=\dfrac 1t \times 1 -0 \times \ln t=\dfrac 1 t$$ The solution to this DE is the vector space $(\ln t , 1)$ Note that for theorem of existence and uniqueness you need to put it in this form: $$y'' +\dfrac {y'}t=0$$