Thread: 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)

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)