Q:

Can you determine a differential equation from its solutions?

A linear first-order differential equation has two solutions:
  $$y_1(x)=x^2 \\y_2(x)=\frac{1}{x}$$ Determine the differential
  equation

I did some research and I think I can use the wronskian to determine my original DE but I dont' really get how it works. Can someone show me how it's done? (It would be nice if you could use a different example so I can solve this question myself).

A:

Let $y'=a(x)\,y+f(x)$ be the equation. We have to determine $a$ and $f$. The function $y_1-y_2$ is a solution of the homogeneous equation $y'=a\,y$, that is
$$
2\,x+\frac{1}{x^2}=a(x)\Bigl(x^2-\frac1x\Bigr).
$$
From this you find $a$, and then $f$.
$$$$$$$$

A:

In case of finding the linear second order differential equation for which $\{y_1,y_2\}$ is the basis of solutions, you can follow your first intuition and solve it using the Wronskian.
First you have to check that $W(y_1,y_2)(x) \ne 0$ for any $x$ on the interval definition of your differential equation.
Hopefully, $\forall x >0 $ or $x<0$ :
$$W(y_1,y_2)(x) =
 \begin{vmatrix}
   y_1 & y_2 \\
   y_1' & y_2' 
\end{vmatrix} 
 =
 \begin{vmatrix}
   x^2 & x^{-1} \\
   2x & -x^{-2}  
\end{vmatrix}  = - 3 \ne 0$$
That means that the following equation
$$  \begin{vmatrix}
  y & y_1 & y_2 \\
   y' & y_1' & y_2' \\  
y'' & y_1'' & y_2''
\end{vmatrix} = 0 = y''  \begin{vmatrix}
   y_1 & y_2 \\
   y_1' & y_2' 
\end{vmatrix} -y'  \begin{vmatrix}
   y_1 & y_2 \\
   y_1'' & y_2'' 
\end{vmatrix} + y  \begin{vmatrix}
   y_1' & y_2' \\
   y_1'' & y_2'' 
\end{vmatrix}  $$
is a true second order differential equation. And you can see that both $y_1$ and $y_2$ are solution to this equation by replacing $y$ by $y_1$ or $y_2$ in the determinant expression.

