Author Topic: TUT0801 QUIZ3  (Read 11 times)

hellocyolo

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TUT0801 QUIZ3
« on: October 12, 2019, 01:26:30 AM »
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).
« Last Edit: October 12, 2019, 01:30:12 AM by hellocyolo »