Author Topic: Problem 2 (morning)  (Read 1397 times)

Victor Ivrii

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Problem 2 (morning)
« on: October 23, 2019, 06:00:35 AM »
(a) Find Wronskian  $W(y_1,y_2)(x)$ of a fundamental set of solutions $y_1(x) , y_2(x)$ for ODE 
\begin{equation*}
x y''-(2x+1)y'+(x+1)y=0.
\end{equation*}
(b) Check that $y_1(x)=e^x$ is a solution and find another linearly independent solution.

(c) Write the general solution, and find solution such that ${y(1)=0, y'(1)=e}$.

Jingjing Cui

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Re: Problem 2 (morning)
« Reply #1 on: October 23, 2019, 07:06:28 AM »
a)
$$
y''-\frac{(2x+1)}{x}y'+\frac{x+1}{x}y=0\\
W=ce^{\int{-p(x)dx}}=ce^{\int{\frac{(2x+1)}{x}dx}}\\
\int{\frac{(2x+1)}{x}dx}=2x+ln|x|\\
W=ce^{2x+ln|x|}=cxe^{2x}\\
$$
« Last Edit: October 23, 2019, 07:16:21 AM by Jingjing Cui »

Jingjing Cui

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Re: Problem 2 (morning)
« Reply #2 on: October 23, 2019, 07:19:53 AM »
b)
$$
y_1(x)=e^x\\
y_1'(x)=y_1''(x)=e^x\\
x(e^x)-(2x+1)e^x+(x+1)e^x=xe^x-2xe^x-e^x+xe^x+e^x=0\\
so \;\;y_1(x)=e^x \;is \;a \;solution\\
\\
take \;c=1, then \;W=xe^{2x}\\
W=det
\begin{vmatrix}
e^x&y_2(x)\\
e^x&y_2'(x)\\
\end{vmatrix}\\
e^xy_2'(x)-e^xy_2(x)=xe^{2x}\\
y_2'(x)-y_2(x)=xe^{x}\\
p(x)=-1\\
\mu=e^{\int{p(x)}dx}=e^{-x}\\
e^{-x}y_2'(x)-e^{-x}y_2(x)=x\\
e^{-x}y_2(x)=\int{x}dx=\frac{1}{2}x^2+c\\
y_2(x)=\frac{1}{2}x^2e^{x} \;(take\;c=0)\\
$$

Jingjing Cui

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Re: Problem 2 (morning)
« Reply #3 on: October 23, 2019, 07:28:41 AM »
c)
$$
y(x)=c_1(e^x)+c_2(\frac{1}{2}x^2e^x)\\
y(1)=c_1(e^1)+c_2(\frac{1}{2}1^2e^1)=c_1e+c_2(\frac{1}{2}e)=0\\
c_1=-\frac{1}{2}c_2\\
y'(x)=c_1(e^x)+c_2xe^x+c_2(\frac{1}{2}x^2e^x)\\
y'(1)=c_1(e^1)+c_2e^1+c_2(\frac{1}{2}1^2e^1)=c_1e+c_2e+\frac{1}{2}c_2e=e\\
c_1+c_2+\frac{1}{2}c_2=1\\
-\frac{1}{2}c_2+c_2+\frac{1}{2}c_2=1\\
c_2=1\\
c_1=-\frac{1}{2}\\
y(x)=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x\\
$$

Mengyuan Wang

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Re: Problem 2 (morning)
« Reply #4 on: October 23, 2019, 07:40:41 AM »
begin{equation}
   \begin{array}{l}{y^{\prime \prime}-\frac{2 x+1}{x} y^{\prime}+(x+1) y=0} \\ {w=c e^{-f(x) d x}=C e^{b+\frac{1}{x} d x}=cx e^{2 x}}\end{array}
   \end{equation}

          let $ C=1 $
      \begin{equation}
         w=\left|\begin{array}{cc}{e^{x}} & {y_{2}} \\ {e^{x}} & {y_{2}^{\prime}}\end{array}\right|=x e^{2 x}
      \end{equation}


\begin{equation}
\begin{array}{c}{e^{x} y^{\prime}-e^{x} y=x e^{2 x}} \\ {y^{\prime}-y=x e^{x}}\end{array}
\end{equation}
\begin{equation}
u=e^{\int-1 d x}=e^{-x}
\end{equation}
\begin{equation}
\begin{aligned} e^{-x} y^{\prime}-e^{-x} y &=x \\ e^{-x} y &=\int x d x \\ e^{-x} y &=\frac{x^{2}}{2}+c \end{aligned}
\end{equation}
let $ C=1 $
\begin{equation}
y_{2}=\frac{x^{2} e^{x}}{2}+e^{x}
\end{equation}
\item[c] so  \begin{equation}
\begin{array}{l}{y=c_{1} e^{x}+c_{2}\left(\frac{x^{2} e^{x}}{2}+e^{x}\right)} \\ {y^{\prime}=c_{1} e^{x}+c_{2}\left(\frac{2 x e^{x}}{2}+\frac{e^{x} x^{2}}{2}+e^{x}\right)}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{y(1)=0} \\ {y^{\prime}(1)=e}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{c_{1} e+G\left(\frac{e}{2}+e\right)=0} \\ {c_{1}+\frac{e}{2} c_{2}=0} \\ {c_{1} e+c_{2}\left(-e+\frac{e}{2}+e\right)= e} \\ {c_{1}+\frac{5}{2} c_{2}=1}\end{array}
\end{equation}
\begin{equation}
\begin{array}{l}{C_{2}=1} \\ {C_{1}=-\frac{3}{2}}\end{array}
\end{equation}
so
\begin{equation}
y=-\frac{3}{2} e^{x}+\left(\frac{x^{2}}{2} e^{-\frac{2}{2}}+e^{x}\right)
\end{equation}

Elenalwaysmiles

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Re: Problem 2 (morning)
« Reply #5 on: October 23, 2019, 08:15:34 AM »
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

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Re: Problem 2 (morning)
« Reply #6 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.

Yiyun Sun

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Re: Problem 2 (morning)
« Reply #7 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, $y = -\frac{1}{2}e^{x}+\frac{1}{2}x^{2}e^{x}$

AllanLi

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Re: Problem 2 (morning)
« Reply #8 on: October 23, 2019, 09:14:42 AM »
\begin{equation}
xy''-(2x+1)y'+(x+1)y=0
\end{equation}
\begin{equation}
y''-\frac{2x+1}{x}y'+\frac{x+1}{x}y=0
\end{equation}p(x)=(2x+1)/x
\begin{equation}
𝝻=e^{-∫-\frac{2x+1}{x}dx}
\end{equation}W=C𝝻
\begin{equation}
𝝻=Ce^{2x}x
\end{equation}W=C𝝻
\begin{equation}
W = Ce^{2x}x
\end{equation}let C =1,then we have
\begin{equation}
W{(y_1,y_2)}=xe^{2x}
\end{equation}for part b), to check that y1 is a solution then
\begin{equation}
y_1' = e^x,y_1'' = e^x
\end{equation}
\begin{equation}
xe^x-(2x+1)e^x+(x+1)e^x=0
\end{equation}so y1 is indeed a solution.
\begin{equation}
W_{(y_1,y_2)}=xe^{2x}=e^xy_2'-e^xy_2
\end{equation}
\begin{equation}
y_2'-y_2=xe^x
\end{equation}the integrating factor is
\begin{equation}
𝝻 = e^{-x}
\end{equation} multiply 𝝻 in this equation
\begin{equation}
e^{-x}y_2'-e^{-x}y_2=x
\end{equation}
\begin{equation}
(e^{-x}y_2)'=x
\end{equation}take integral on both side
\begin{equation}
e^{-x}y_2=\frac{1}{2}x^2
\end{equation}move all the x to one side
\begin{equation}
y_2 = \frac{1}{2}x^2e^x
\end{equation}part c) , the general solution is
\begin{equation}
W = C_1y_1+C_2y_2
\end{equation}
\begin{equation}
y = C_1e^x+\frac{C_2}{2}x^2e^x
\end{equation}
\begin{equation}
y' = C_1e^x+\frac{C_2}{2}(2xe^x+x^2e^x)
\end{equation}we have y(1) = 0 and y(1)'=e
\begin{equation}
C_1e+\frac{C_2}{2}e=0, C_1e+\frac{C_2}{2}(2e+e)=e
\end{equation}solve for C1 and C2,then we have
\begin{equation}
C_1 =-\frac{1}{2},C_2 = 1
\end{equation}
\begin{equation}
y=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x
\end{equation}

GuangyuDu

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Re: Problem 2 (morning)
« Reply #9 on: October 23, 2019, 10:05:07 AM »
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 $y=-\frac12e^x+\frac12x^2e^x$.

lyujiahe

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Re: Problem 2 (morning)
« Reply #10 on: October 28, 2019, 06:03:16 PM »
Do we need to consider (or simply analyze) the situation x = 0? Since all solutions here are based on x=0.