A [not very] Interesting Dilemma

We have moved!

We want to extend a heartfelt thanks to everyone who has made the original ATAR Notes forum such a helpful, warm and welcoming place. Your contributions are appreciated and we will be leaving the forum in archive mode for posterity.

Please visit our new website and discussion area.

Welcome, Guest. Please login or register.

Pages: [1] Go Down

Author Topic: A [not very] Interesting Dilemma (Read 1007 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

Mao

CH41RMN
Honorary Moderator
Great Wonder of ATAR Notes
Posts: 9181
Respect: +390
School: Kambrya College
School Grad Year: 2008

A [not very] Interesting Dilemma

« on: November 08, 2009, 11:44:16 pm »

Came across this problem in the textbook, and I cannot seem to arrive at the correct answer. Mathematica confirms my workings are correct, which must mean the method breaks down:

Solve $x^2 \frac{d^2 y}{dx^2} -2x \frac{dy}{dx} + 2y = x^3 \cos x$ using the method of variation of parameters

$y_h(x) = c_1 x^2 + c_2 x$ (homogenous solution to the DE)

$y_p(x) = y_1 \int \frac{-y_2 \cdot r}{W}\; dx + y_2 \int \frac{y_1 \cdot r}{W}\; dx$ , where $W = \left| \begin{array}{cc} y_1 & y_2 \\ y_1' & y_2' \\ \end{array} \right|$

After much workings

$y_p(x) = 6 x \cos(x) - x^3 \cos(x) + 4 x^2 \sin(x)$ , which when substituted back into the DE, gives $x^5 \cos x$

Would someone like to please point me in the right direction or tell me what is wrong?

Cheers,

The actual answer is $y_p(x) = -x\cos (x)$ , which is verified.

Logged

Editor for ATARNotes Chemistry study guides.

VCE 2008 | Monash BSc (Chem., Appl. Math.) 2009-2011 | UoM BScHon (Chem.) 2012 | UoM PhD (Chem.) 2013-2015

Search the forums now!

We have moved!

Author Topic: A [not very] Interesting Dilemma (Read 1007 times) Tweet Share

Mao

A [not very] Interesting Dilemma

Recent Posts

User:
Password: