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### Messages - Yiyun Sun

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##### Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 08:23:20 AM »
I use reduction of order to determine y2. This is my solution.

(a)Rewrite the equation:
$y'' - \frac{2x+1}{x}y' + \frac{x-1}{x}y = 0$
Then $p(x) = -\frac{2x+1}{x} = -(2 +\frac{1}{x})$
By Abel's Theorem, we have
$W(y_1, y_2)(x) = c\ exp(\int-p(x)dx) = c\ exp(\int2 +\frac{1}{x}dx) = cxe^{2x}$
Let c = 1,
$W(y_1, y_2)(x) = xe^{2x}$

(b)Since $y_1(x) = e^{x}$, so $y_1'(x) = e^{x}$ and $y_1''(x) = e^{x}$
Plug in :
$xe^{x} - (2x+1)e^{x} + (x-1)e^{x}$
$=xe^{x} - 2xe^{x} + e^{x} + xe^{x} - e^{x}$
$=0$
So $y = e^{x}$ is a solution.

By reduction of order, we have,
$y_2 = y_1 \int(\frac{e^{\int-p(x)dx}}{y_1^{2}})dx =e^{x} \int(\frac{xe^{2x}}{e^{2x}})dx =e^{x} \int(x)dx =e^{x}(\frac{1}{2}x^{2} + C)$
Let C = 0,
$y_2 = \frac{1}{2}x^{2}e^{x}$

(c)The general solution is:
$y = c_1e^{x} + c_2\frac{1}{2}x^{2}e^{x}$
Then $y' = c_1e^{x} + c_2\frac{1}{2}x^{2}e^{x} +c_2xe^{x}$
Since $y(1) = 0, y'(1) = e$,
So, $ec_1 + \frac{1}{2}ec_2 = 0$ and $ec_1 + \frac{1}{2}ec_2 + ec_2 = e$
Thus, $c_1 = -\frac{1}{2} , c_2 = 1$
Therefore, $y = -\frac{1}{2}e^{x}+\frac{1}{2}x^{2}e^{x}$

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##### Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 08:21:18 AM »
Elenalwaysmiles, I think Wang's answer is correct. It's okay to take a different C for the solution y2. And if you rearrange Wang's final answer, you will get same answer with your answer where c1 = -1/2 and c2 = 1/2.

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