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### Messages - xilin zhang

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1
##### Quiz-7 / QUIZ7 LEC0201
« on: November 29, 2019, 02:21:27 PM »
Solutions in the attachments. 2
##### Term Test 2 / Re: Problem 1 (noon)
« on: November 19, 2019, 09:11:19 AM »
I got a different y' in part b.

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##### Quiz-6 / QUIZ6 LEC0201
« on: November 15, 2019, 06:03:27 PM »
Please see solution in the attachement.

4
##### Quiz-4 / TUT0801 QUIZ4
« on: October 19, 2019, 12:02:53 AM »
Complete solution is attached.

5
##### Quiz-3 / 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).

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