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# FP1 - Second Order Differential Equation watch

1. Hi,

I can't seem to get the correct answer when solving this differential equation:

4d²y/dx² - 4dy/dx + 5y = 0

as the answer is e^(1/2x)[Acosx+Bsinx], although I am getting
e^x[Acos(x/2)+Bsin(x/2)]

Thanks!
2. Hmm shouldnt give out full answers.

Could you provide us with your workings out?
3. 4d²y/dx² - 4dy/dx + 5y = 0

Auxiliary equation:

4m²-4m+5=0
(2m-2)²-4+5=0
(2m-2)²=-1
2m-2=±i
m=1±(1/2)i

therefore, the general solution is:

y=Pe^[1+(1/2)i]x + Qe^[1-(1/2)i]x
y=e^x{Pe^[(1/2)ix] + Qe^[(-1/2)ix]}
y=e^x(Acos[(1/2)x] + Bcos[(1/2)x]) where A=P+Q and B=i(P-Q)
4. Now that is weird. I used the quadratic equation for this question.

That will give you the right answer.

However i dont understand why completing the square didnt work .
5. 4m² - 4m + 5 = 0
(2m - 2)² - 4 + 5 = 0

Checking this...

(2m - 2)(2m - 2) - 4 + 5 = 0
4m^2 - 4m - 4m + 4 - 4 + 5 = 0
4m^2 - 8m + 5 = 0

Not what you wanted.
6. aahhhh damn should've seen that.
7. Ohh I see, I've completed the square wrong! I'll have a go at that again, thanks for your help everyone!

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Updated: October 14, 2006
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