Verify that the function y1 and y2 are solutions of the given differential equation. Do $y1$ and $y2$ constitute a fundamental set of solutions?
Considering the differential equation:
$x^2y''-x(x+2)y'+(x+2)y=0$ (1)
Use direct substitution, to verify that $y1(x)=e^t$ is a solution to the differential equation.
Because $y1(x)=x,y'=1,y''=0$, substitute these into equation (1):
we can verify that the left hand side equals 0. Thus, we know y1(x) is a solution for equation (1).
Next, we want to see if y2(x)=$xe^x$ satisfies equation (1): we know that ,$y2(x)'=xe^x+e^x$, $y2(x)''=e^x(x+2)$. Then, we can see that is a solution for equation(1). After substitution, we see this is a solution for equation(1).
Recall the Wronskian of the functions, y1(t) and y2(t):
$W(y1,y2) = det
\begin{vmatrix}
y1(x)& y2(x)\\
y1'(x) & y2'(x)
\end{vmatrix}
=x(e^x+e^x)-xe^x
=x^2e^x$
since $x>0$ and $e^x>0$, $W(y1,y2)>0$,
Therefore, y1(x),y2(x) is a fundamental set of equation (1).
