Question

ODE/algebra question http://i.imgur.com/3sMDW.png

Answer 1

I currently stuck on b(ii) on this. From what I understand I sub y1 into the original equation. I do this by derving y1 twice using the product rule.

Answer 2

I have done the computation here, I'll attached it below

Answer 3

I've done a question like this previously, it should all cancel out, not here it seems.

Answer 4

Sorry, this should work http://postimage.org/image/lelva97fx/

Answer 5

If y1 is a particular solution then
\[u(x).y_1(x)\] will be a solution for some u(x).
Put
\[y(x)=u(x)y_1(x)\]
\[y'(x)=u'(x)y_1(x)+y_1'(x)u(x)\]
\[y''(x)=u''(x)y_1(x)+y_1''(x)u(x)+2y_1'(x)u(x)\]
use \[y_1=e^{x/3}\]
solve for u(x)

Answer 6

That's what I did, it didn't solve!

Answer 7

well may be u didn't do it right ur calculations are messy it is easy for u to get stuck

Answer 8

I am pretty sure I did it right, I went through it slowly.

Answer 9

I can't see any calculation errors anyways.

Answer 10

So did I got u"=0

Answer 11

Where did I go wrong??

Answer 12

I've being literally staring at my page for the past 40mins looking for an error.

Answer 13

Sorry i don't have a scanner

Answer 14

Have you seen my work? Do you see an error? http://postimage.org/image/lelva97fx/

Answer 15

Wait i rechecked my calculation u are correct the question contains an error.u" cannot be zero

Answer 16

my bad correction

Answer 17

no sorry question is correct

Answer 18

yeah it is correct
Checked it three times

Answer 19

Can you see where I went wrong?

Answer 20

Do it again neatly on a clean sheet of paper
i cannot makeout anything out of ur calculations

Answer 21

I don't see the problem the very last line on my sheet is after I multiply everything out.