40xy' -ln(x^8) = 0

Initial Condition: y(1)=31

Work:

40x (dy/dx) - ln(x^8) = 0

40x(dy/dx) = ln (x^8)

dy/dx = ln(x^8) / 40x

Sdy = Sln(x^8) / 40x

And then...stuck (if its even right thus far)

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- January 23rd 2011, 09:29 AMBooGTSParticular Solution
40xy' -ln(x^8) = 0

Initial Condition: y(1)=31

Work:

40x (dy/dx) - ln(x^8) = 0

40x(dy/dx) = ln (x^8)

dy/dx = ln(x^8) / 40x

Sdy = Sln(x^8) / 40x

And then...stuck (if its even right thus far) - January 23rd 2011, 09:35 AMChris L T521
- January 23rd 2011, 09:36 AMchisigma
Because is the DE becomes...

(1)

... where the variables are separated so that the solution is immediate...

Kind regards

- January 23rd 2011, 09:49 AMHallsofIvy
- January 23rd 2011, 01:42 PMBooGTS
Still not doing this right I don't think:

y= 1/5 (lnxdx)/x

= 1/5 u du

= 1/5 u^2/2

= 1/10 lnx^2 - January 23rd 2011, 02:58 PMskeeter
- January 23rd 2011, 03:42 PMBooGTS
Yeah, big difference. Thanks to all (and on the site if I can figure out how)