# Particular Solution

• Jan 23rd 2011, 08:29 AM
BooGTS
Particular 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 AM
Chris L T521
Quote:

Originally Posted by BooGTS
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)

Hint: $\displaystyle \ln(x^8)=8 \ln x$. Then the integral on the RHS becomes $\displaystyle \displaystyle\frac{1}{5}\int\frac{\ln x}{x}\,dx$.

Do you know how to proceed?

(And yes, everything is right so far. (Nod) )
• Jan 23rd 2011, 08:36 AM
chisigma
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 AM
HallsofIvy
Quote:

Originally Posted by Chris L T521
Hint: $\displaystyle \ln(x^8)=8 \ln x$. Then the integral on the RHS becomes $\displaystyle \displaystyle\frac{1}{5}\int\frac{\ln x}{x}\,dx$.

Do you know how to proceed?

(And yes, everything is right so far. (Nod) )

If you are concerned about how to do that integral, let u= ln(x).
• Jan 23rd 2011, 12:42 PM
BooGTS
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 PM
skeeter
Quote:

Originally Posted by BooGTS
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

note the difference between $\displaystyle \ln{x^2}$ and $\displaystyle (\ln{x})^2$ ...

the correct antiderivative is

$\displaystyle \displaystyle \frac{(\ln{x})^2}{10} + C$
• Jan 23rd 2011, 02:42 PM
BooGTS
Yeah, big difference. Thanks to all (and on the site if I can figure out how)