Author Topic: WW3 -Q13  (Read 2725 times)

Boon Yuen Tan

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WW3 -Q13
« on: October 01, 2014, 08:23:15 PM »
Any idea how to solve this?
$$x^2y′′−9xy′+25y=x^9, \qquad   y(1)=7,\quad  y′(1)=0. $$

I can't use Wronskian to solve for $y_2$  cause $y_1$ was not given
« Last Edit: October 07, 2014, 03:23:24 PM by Victor Ivrii »

Victor Ivrii

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Re: WW3 -Q13
« Reply #1 on: October 02, 2014, 04:13:56 AM »
It is Euler equation. See Handout 4b and the textbook problem 34 on page 166.

Yeming Wen

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Re: WW3 -Q13
« Reply #2 on: October 07, 2014, 02:16:00 PM »
Hi, prof,

Just wondering whether it is possible to guess that Ax^9 is a particular solution and then we can solve that A=1/16.
Final, we use reduction of order to find the general solution.

Victor Ivrii

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Re: WW3 -Q13
« Reply #3 on: October 07, 2014, 03:25:02 PM »
Hi, prof,

Just wondering whether it is possible to guess that $Ax^9$ is a particular solution and then we can solve that $A=1/16$.
Indeed
Quote
Final, we use reduction of order to find the general solution.
Would not work. You need to read handout 4b!

Christopher Choi

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Re: WW3 -Q13
« Reply #4 on: October 07, 2014, 07:06:22 PM »
You have to first find out the answer
x2y''−9xy'+25y=0
then that would be your Yc.

Then you need to use the method , variation of parameters from 3.6 to find Yp.

Your final answer would be Y=Yc+Yp

Victor Ivrii

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Re: WW3 -Q13
« Reply #5 on: October 07, 2014, 07:31:48 PM »
Then you need to use the method , variation of parameters from 3.6 to find Yp.
No need. Guess $y=Ax^9$ is good (because $9$ is not a characteristic root).