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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- Jan 23rd 2011, 08:29 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) - Jan 23rd 2011, 08:35 AMChris L T521
- Jan 23rd 2011, 08:36 AMchisigma
Because is $\displaystyle \ln x^{8} = 8\ \ln x$ the DE becomes...

$\displaystyle \displaystyle y^{'} = \frac{\ln x}{5 x}$ (1)

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jan 23rd 2011, 08:49 AMHallsofIvy
- Jan 23rd 2011, 12:42 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 - Jan 23rd 2011, 01:58 PMskeeter
- Jan 23rd 2011, 02:42 PMBooGTS
Yeah, big difference. Thanks to all (and on the site if I can figure out how)