### Author Topic: TT1-problem 2  (Read 11578 times)

#### Victor Ivrii

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##### TT1-problem 2
« on: October 09, 2014, 01:59:21 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^3(\ln x+1)\cdot  y''(x) -(2\ln x+3)x^2 \cdot y'(x) + (2\ln x+3) x y(x) = 0,\qquad
x>1.
\end{equation*}

b. Check that $y_1(x)=x$ is a solution and find another linearly independent solution.

#### Yeming Wen

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##### Re: TT1-problem 2
« Reply #1 on: October 09, 2014, 11:13:46 AM »
Part a)
First we make the coefficient of $y''$ one, which results in
\begin{equation*}
y''(x) - \frac{(2\ln x+3)}{x(\ln x+1)} y'(x) + \frac{(2\ln x+3)}{x^2(\ln x+1)}y(x)=0
\end{equation*}
By Abel' Thm,

\ln W=\int{\frac{(2\ln x+3)}{x(\ln x+1)}}dx\label{A}

Using change of variables to evaluate the integral,let
\begin{equation*}
\mu=\ln x + 1
\end{equation*}
The the integral becomes
\begin{equation*}
\int{\frac{2\mu+1}{\mu}}du =\int{2}du+\int{\frac{1}{\mu}}du \label{B}
\end{equation*}
We have \begin{equation*} \ln W=(2\mu+\ln \mu)+C_1   \end{equation*}
Plug in x back, which is $$W= C_2x^2(\ln x + 1)$$

Part b)
To verify y = x is a solution. Just plug in, we have
\begin{equation*} x^3(\ln x+1)\cdot  0 -(2\ln x+3)x^2 \cdot 1 + (2\ln x+3) x^2  = 0 \end{equation*}
Then to find another independent solution, notice that $W > 0$ for $x > 1$
So we can just let $C_2 = 1$. And we know $W = \left| \begin{matrix} x & y \\1 & yâ€™ \end{matrix}\right|$
Now we have a new ode
$$y' - \frac{1}{x} y = x(\ln x + 1), \qquad x>1.$$
Solve the homogeneous ode first,
\begin{equation*} z' - \frac{1}{x}z = 0 \end{equation*}
We get $z=x$. Then we know $y=\mu(x)x$.

And $y'=\mu'x+\mu$, we plug in into (3). We have \begin{equation*} u'=(\ln x + 1) \end{equation*}
Then \begin{equation*} u=\int{\ln x +1}dx \end{equation*}
We use integral by parts to evaluate $\int{\ln x}dx$ first.
\begin{equation*} \int{\ln x} dx = x\ln x - \int{1}dx = x\ln x - 1 \end{equation*}
So we have \begin{equation*} \mu x=\ln x + x + C \end{equation*}
Finally, we have $y_2=(x\ln x+x+C)x$.
« Last Edit: October 09, 2014, 03:38:58 PM by Yeming Wen »

#### Victor Ivrii

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##### Re: TT1-problem 2
« Reply #2 on: October 09, 2014, 11:51:29 AM »
Obviously you use the wrong formula for $W$ in (\ref{A}) and miss $d\mu$ in (\ref{B}). Fix!

#### Roro Sihui Yap

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##### Re: TT1-problem 2
« Reply #3 on: October 09, 2014, 12:28:39 PM »
\begin{equation*}
y''(x) - \frac{(2\ln x+3)}{x(\ln x+1)} y'(x) + \frac{(2\ln x+3)}{x^2(\ln x+1)}y(x)=0
\end{equation*}
Using Abel's Theorem,

W=ce^{\int{\frac{(2\ln x+3)}{x(\ln x+1)}}}dt

\int{\frac{(2\ln x+3)}{x(\ln x+1)}dt}  = \int{\frac{(2\ln x+4)}{x(\ln x+1)}dt} - \int{\frac{1}{x(\ln x+1)}dt} \\
\int{\frac{(2\ln x+3)}{x(\ln x+1)}dt}  = 2ln(xln(x) + x) - ln(ln(x) +1) \\
\int{\frac{(2\ln x+3)}{x(\ln x+1)}dt}  = ln(x^2 (lnx +1))

Therefore,

W = c(x^2 (ln(x) +1))

#### Roro Sihui Yap

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##### Re: TT1-problem 2
« Reply #4 on: October 09, 2014, 12:32:32 PM »
2b
$y_1(x)=x$, $y_1'(x)=1$, $y_1''(x)=0$. Substituting into the equation we get,

x^3(\ln x+1)\cdot  (0) -(2\ln x+3)x^2   + (2\ln x+3) x^2 = 0

$y_1(x)=x$ is indeed a solution. We know that

W = c(x^2 (\ln(x) +1))

and also  $W = \det \begin{bmatrix} x & g(t) \\1 & gâ€™(t)\end{bmatrix}$  or simply $W = \left| \begin{matrix} x & g(t) \\1 & gâ€™(t)\end{matrix}\right|$

So,

xg'(t) - g(t) = c(x^2 (ln(x) +1))\\
g'(t) - g(t)/x = c(xln(x) +x)

Find an integrating factor  $d\mu/dx = e^{-\int1/x}$. $\mu = x^{-1}$. Multiply by $x^{-1}$. Throughout the equation  $g'(t)/x - g(t)/x^2 = c(\ln(x) + 1)$, $[g(t)/x]' = c\ln x + c$, $g(t)/x = cx\ln x - cx + cx \implies g(t) = cx^2\ln x$ -- The second solution .
« Last Edit: October 09, 2014, 02:43:47 PM by Victor Ivrii »

#### Bogdan Scaunasu

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##### Re: TT1-problem 2
« Reply #5 on: October 09, 2014, 12:44:36 PM »
2a) Changing the equation to:
$$y'' - \frac{2 \ln x + 3}{x (\ln x + 1)} y' + \frac{2 \ln x + 3}{x ^ 2 (\ln x + 1)} y = 0$$
Let $-p = \frac{2 \ln x + 3}{x (\ln x + 1)}$. Then:
\begin{aligned}
-\int p dx & = \int \frac{2 \ln x + 3}{x (\ln x + 1)} dx \\
& = \int \frac{2 \ln x + 2}{x (\ln x + 1)} dx + \int \frac{1}{x (\ln x + 1)} dx \\
& = 2 \int \frac{1}{x} dx + (\ln(\ln(x) + 1) + \ln C) \\
-\int p dx & = 2 \ln x + \ln(\ln(x) + 1) + \ln C
\end{aligned}
Using Abel's Theorem:
$$W = c e ^ {- \int p dx}$$
\begin{aligned}
W & = c e ^ {- \int p dx} \\
& = c e ^ {2 \ln x + \ln(\ln(x) + 1) + \ln C} \\
& = c (C x ^ {2} (\ln(x) + 1))
\end{aligned}
Therefore:
$$\boxed{W = C x ^ {2} (\ln(x) + 1)}$$

** fixed minus sign
« Last Edit: October 09, 2014, 04:43:33 PM by Bogdan Scaunasu »

#### Victor Ivrii

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##### Re: TT1-problem 2
« Reply #6 on: October 09, 2014, 02:45:35 PM »
Guys, there is no point to post solution which has been posted already by someone else (especially, as Bogdan didâ€”-with an error)

#### Bogdan Scaunasu

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##### Re: TT1-problem 2
« Reply #7 on: October 09, 2014, 04:23:33 PM »
I apologize for posting a solution after someone else did. I was writing mine and did not notice.
« Last Edit: October 09, 2014, 04:28:11 PM by Bogdan Scaunasu »

#### Sang Wu

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##### Re: TT1-problem 2
« Reply #8 on: October 09, 2014, 08:32:54 PM »
a. W = x2(lnx + 1)
b. W= xy2' - y2 = x2(lnx + 1)
u = e^(âˆ«-1/x * dx) = x^(-1)
(y2/x)' = lnx + 1
y2/x = âˆ«(lnx + 1)dx = âˆ«lnxdx + x = xlnx - x + x = xlnx
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