### Author Topic: MT problem 3  (Read 6839 times)

#### Victor Ivrii

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##### MT problem 3
« on: October 29, 2014, 08:59:21 PM »
Find the general solution of
\begin{equation*}
x^3 y''' + 6x^2 y'' + 5x y' -5y = x^2 \ln x \ .
\end{equation*}
Hint: Note that $1$ is the root of the corresponding characteristic polynomial.

#### Roro Sihui Yap

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##### Re: MT problem 3
« Reply #1 on: October 30, 2014, 01:25:34 AM »
Let $t = ln x$
$x =e^t$

Let $\frac{dy}{dx} = e^{-t}\frac{dy}{dt}$
Let $\frac{d^2y}{dx^2} = e^{-2t}\frac{d^2y}{dt^2} - e^{-2t}\frac{dy}{dt}$
Let $\frac{d^3y}{dx^3} = e^{-3t}\frac{d^3y}{dt^3} - 3e^{-3t}\frac{d^2y}{dt^2} + 2e^{-3t}\frac{dy}{dt}$

$x^3 y''' + 6x^2 y'' + 5x y' -5y = x^2 \ln x \$
$e^{3t} (e^{-3t}\frac{d^3y}{dt^3} - 3e^{-3t}\frac{d^2y}{dt^2} + 2e^{-3t}\frac{dy}{dt}) + 6e^{2t} ( e^{-2t}\frac{d^2y}{dt^2} - e^{-2t}\frac{dy}{dt}) + 5e^t (e^{-t}\frac{dy}{dt}) - 5y = te^{2t}$
$\frac{d^3y}{dt^3} - 3\frac{d^2y}{dt^2} + 2\frac{dy}{dt}+ 6\frac{d^2y}{dt^2} - 6\frac{dy}{dt}+ 5\frac{dy}{dt} - 5y = te^{2t}$
$\frac{d^3y}{dt^3} + 3\frac{d^2y}{dt^2} + \frac{dy}{dt} - 5y = te^{2t}$
\begin{gather}  y''' + 3y'' + y' - 5y = te^{2t} \end{gather}

Find the solution to the homogeneous equation,
$r^3 + 3r^2 + r - 5 = 0$
$(r - 1)(r^2 + 4r + 5) = 0$
r = 1, -2 - i, -2 + i

The solution to the homogeneous equation is
$y = c_1e^t + c_2e^{-2t}cos(t) + c_3e^{-2t}sin(t)$
$y = c_1x + \frac{c_2}{x^2}cos(lnx) + \frac{c_3}{x^2}sin(lnx)$

Use equation 1 and the method of undetermined coefficients to find the non homogeneous solution
let $y_p = Ate^{2t} + Be^{2t}$
$y_p' = Ae^{2t} + 2Ate^{2t} + 2Be^{2t}$
$y_p'' = 4Ae^{2t} + 4Ate^{2t} + 4Be^{2t}$
$y_p''' = 12Ae^{2t} + 8Ate^{2t} + 8Be^{2t}$

Substitute into equation 1
$12Ae^{2t} + 8Ate^{2t} + 8Be^{2t} + 12Ae^{2t} + 12Ate^{2t} + 12Be^{2t} + Ae^{2t} + 2Ate^{2t} + 2Be^{2t} - 5Ate^{2t} - 5Be^{2t} = te^{2t}$
$8Ate^{2t} + 12Ate^{2t} + 2Ate^{2t} - 5Ate^{2t} = te^{2t}$
$17A = 1$
Therefore A = $\frac{1}{17}$
$12Ae^{2t} + 8Be^{2t} + 12Ae^{2t} + 12Be^{2t} + Ae^{2t} + 2Be^{2t} - 5Be^{2t} = 0$
$25A + 17B = 0$
Therefore B = $\frac{-25}{289}$

$y_p = \frac{1}{17}te^{2t} - \frac{25}{289}e^{2t}$
$y_p = \frac{x^2(ln x)}{17} - \frac{25x^2}{289}$

Therefore $y = c_1x + \frac{c_2}{x^2}cos(lnx) + \frac{c_3}{x^2}sin(lnx) + \frac{x^2(ln x)}{17} - \frac{25x^2}{289}$

#### Yeming Wen

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##### Re: MT problem 3
« Reply #2 on: October 30, 2014, 01:31:46 AM »
$$x^3 y''' + 6x^2 y'' + 5x y' -5y = x^2 \ln x \label{A}$$
First we notice that it is Euler equation. We let \begin{equation*} t = \ ln x \end{equation*}
Then Let D denote the operator, which is \begin{equation*} D = \frac{d}{dt} \end{equation*}
Plug in the equation (\ref{A}), we get
\begin{equation*}  D(D-1)(D-2)y + 6D(D-1)y + 5Dy -5y =  e^{2t} \end{equation*}
Expend it, we can have,
\begin{equation*}  D^3y +  3D^2y + Dy - 5y = e^{2t} \end{equation*}
which is,
$$y''' + 3y'' + y' - 5y = e^{-2t} \label{B}$$
Now we have to solve (\ref{B}), which is a constant third order ODE.
First we solve the homogeneous one, which is
$$y''' + 3y'' + y' - 5y = 0 \label{C}$$
The characteristic equation is
\begin{equation*} r^{3} + 3r^2 + r - 5 = (r - 1)(r^2 + 4r + 5) = 0 \end{equation*}
So the solution to (\ref{C}) is
\begin{equation*} y = c_1e^{t} + e^{-2t}(c_2 \cos(t) + c_3 \sin(t)) \end{equation*}
Now it remains to find a particular solution to (\ref{B}).
Since 2 is not a root of the corresponding characteristic polynomial, then we can guess $y=Ae^{2t}$
Plug in (\ref{B}), we have $A = \frac{1}{17}$.
So a general solution to (\ref{B}) is
$$y = c_1e^{t} + e^{-2t}(c_2 \cos(t) + c_3 \sin(t)) + \frac{1}{17}e^{2t} \label{D}$$
Substitute $t = \ lnx$
We have
$$y = C_1x + x^{-2}(C_2 \cos(\ ln x) + C_3 \sin(\ ln x)) + \frac{1}{17}x^2$$

#### Victor Ivrii

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##### Re: MT problem 3
« Reply #3 on: October 30, 2014, 06:48:41 AM »
Roro is correct, Yeming made an error: there must be $x^2\ln x$ term in the solution (with some coefficient)

Type
Code: [Select]
\ln x , \cos x, \sin ,\exp, \tan  etc

This prints these and other functions upright , not italics ("operators" --not in mathematical but in typographical sense ) must be upright and provides proper horizontal spacing

#### Yuan Bian

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##### Re: MT problem 3
« Reply #4 on: November 18, 2014, 04:47:25 PM »
hi, prof
In the official soln of q3, I think there is a tiny mistake:
First, we set yp = Ax2 ln x + B
but after finding A and B, yp =(1/17)x2 ln x - (25/289)x2....

#### Victor Ivrii

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##### Re: MT problem 3
« Reply #5 on: November 18, 2014, 05:22:14 PM »
In the official soln of q3, I think there is a tiny mistake:
First, we set yp = Ax2 ln x + B

Yes, it should be $y_p=Ax^2\ln (x)+Bx^2$

#### Chang Peng (Eddie) Liu

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##### Re: MT problem 3
« Reply #6 on: December 05, 2014, 11:57:12 PM »
can someone please tell me what section of the textbook this question was from?