Stuck Solving This Differential Equation?

[Laney5]

Im looking for a general solution to the following equation but i can't seem to get an answer.
3xdy = (ln(y^{6}) - 6lnx)ydx

I've got this far anyway..

\frac{3x}{y} dy = (ln(y^{6})-ln(x^{6})) dx

\frac{dy}{dx} = \frac{y}{3x} . 6ln\frac{y}{x}

\frac{dy}{dx} = \frac{y}{x} . 2ln\frac{y}{x}

Let z = \frac{y}{x}

\frac{dz}{dx} = (x\frac{dy}{dx} - y) / x^{2}

\frac{dz}{dx} = \frac{1}{x}(\frac{y}{x} . 2ln\frac{y}{x}) - \frac{y}{x^2}

z = \frac{y}{x} gives

\frac{dz}{dx} = \frac{1}{x}(2zlnz) - \frac{z}{x}

\frac{dz}{dx} = \frac{1}{x}(2zlnz-z)

\frac{dz}{2zlnz - z} = \frac{1}{x}dx

now I am stuck...??

 

[HallsofIvy]

I presume it is the
\frac{dz}{2zlnz- z}
that is giving you the problem!

Write it as
\frac{1}{2ln z- 1}\frac{dz}{z}
and let u= 2ln z- 1.

 

[Laney5]

Ya, its that part alright!

Do i write it like \frac{1}{u} \int \frac{dz}{z}?

\frac{1}{u} \int \frac{dz}{z} = Integral(\frac{dx}{x})



\frac{1}{u} (lnz) = lnx + C

\frac{1}{2lnz - 1} (lnz) = lnx + C

Is this right?

 

[Hootenanny]

Ya, its that part alright!

Do i write it like \frac{1}{u} \int \frac{dz}{z}?

\frac{1}{u} \int \frac{dz}{z} = Integral(\frac{dx}{x})



\frac{1}{u} (lnz) = lnx + C

\frac{1}{2lnz - 1} (lnz) = lnx + C

Is this right?

No. Notice that u is a function of z and is therefore not constant under integration with respect to z. Therefore you need to make a change of variable from dz\mapsto du.

 

[Laney5]

u=2lnz-1
du=2(\frac{1}{z})dz
(\frac{1}{2})du=(\frac{1}{z})dx

(\frac{1}{u}).(\frac{1}{2})du=(\frac{1}{x})dx

\frac{1}{2}\int\frac{1}{u}du=\int\frac{1}{x}dx

(\frac{1}{2})lnu=lnx + C

ln(u)^{1/2} - lnx = C

ln[\frac{(u)^{1/2}}{x}] = C

u=2lnz-1

ln[\frac{(2lnz-1)^{1/2}}{x}] = C

z=\frac{y}{x}

ln[\frac{(2ln\frac{y}{x}-1)^{1/2}}{x}] = C

e^{ln[\frac{(2ln\frac{y}{x}-1)^{1/2}}{x}] } = e^C , Let e^C = B

\frac{(2ln\frac{y}{x}-1)^{1/2}}{x} = B

Am i doing it correctly?

 

[Hootenanny]

Looks okay to me