MAT244--2019F > Term Test 1

Problem 2 (morning)

<< < (2/3) > >>

Elenalwaysmiles:
Mengyuan Wang, I think you made a mistake.
The second linear independent solution y2 should be just (x^2e^x)/2. I think it is caused by the additional c. If you do not include the c when calculating y2, your results will be right.
And the coefficients of the general solution are also wrong, c1 should be -1/2, c2 should be 1/2.

Yiyun Sun:
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.

Yiyun Sun:
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, $y = -\frac{1}{2}e^{x}+\frac{1}{2}x^{2}e^{x}$

AllanLi:

xy''-(2x+1)y'+(x+1)y=0

y''-\frac{2x+1}{x}y'+\frac{x+1}{x}y=0
p(x)=(2x+1)/x

𝝻=e^{-∫-\frac{2x+1}{x}dx}
W=C𝝻

𝝻=Ce^{2x}x
W=C𝝻

W = Ce^{2x}x
let C =1,then we have

W{(y_1,y_2)}=xe^{2x}
for part b), to check that y1 is a solution then

y_1' = e^x,y_1'' = e^x

xe^x-(2x+1)e^x+(x+1)e^x=0
so y1 is indeed a solution.

W_{(y_1,y_2)}=xe^{2x}=e^xy_2'-e^xy_2

y_2'-y_2=xe^x
the integrating factor is

𝝻 = e^{-x}
multiply 𝝻 in this equation

e^{-x}y_2'-e^{-x}y_2=x

(e^{-x}y_2)'=x
take integral on both side

e^{-x}y_2=\frac{1}{2}x^2
move all the x to one side

y_2 = \frac{1}{2}x^2e^x
part c) , the general solution is

W = C_1y_1+C_2y_2

y = C_1e^x+\frac{C_2}{2}x^2e^x

y' = C_1e^x+\frac{C_2}{2}(2xe^x+x^2e^x)
we have y(1) = 0 and y(1)'=e

C_1e+\frac{C_2}{2}e=0, C_1e+\frac{C_2}{2}(2e+e)=e
solve for C1 and C2,then we have

C_1 =-\frac{1}{2},C_2 = 1

y=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x

GuangyuDu:
Question 2:
$xy''-(2x+1)y'+(x+1)y=0$

Find wronskian form and $y_2$, given $y_1(x)=e^x$.
$y(1)=0,y'(1)=e$

Solution:
$y''-\frac{2x+1}{x}y'+\frac{x+1}{x}y=0$
$w=ce^{-\int p(x)\mathrm{d}x}=ce^{\int \frac{2x+1}{x}\mathrm{d}x}=ce^{2x}e^{\ln x}$

Let $c=1$, $w=xe^{2x}$
$\begin{bmatrix} y_1&y_2\\ y_1'&y_2'\\ \end{bmatrix} = \begin{bmatrix} e^x&y_2\\ e^x&y_2'\\ \end{bmatrix} = e^xy_2'-e^xy_2=xe^{2x}$
$y_2'-y_2=xe^x$
$M=e^{\int-1\mathrm{d}x }=e^{-x}$

Multiple $e^{-x}$ on both sides.
$e^{-x}y_2'-e^{-x}y_2=x$
$(e^{-x}y_2)'=x$
$y_2=\frac12x^2e^x+e^x$
$y=c_1y_1+c_2y_2=c_1e^x+c_2\left(\frac12x^2e^x+e^x\right)$

plug in $y(1)=0,y'(1)=e$.

we have $y=-\frac12e^x+\frac12x^2e^x$.