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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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: . Then the integral on the RHS becomes . Do you know how to proceed? (And yes, everything is right so far. )
Because is the DE becomes... (1) ... where the variables are separated so that the solution is immediate... Kind regards
Originally Posted by Chris L T521 Hint: . Then the integral on the RHS becomes . Do you know how to proceed? (And yes, everything is right so far. ) If you are concerned about how to do that integral, let u= ln(x).
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
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 and ... the correct antiderivative is
Yeah, big difference. Thanks to all (and on the site if I can figure out how)
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