Cant find the general solution

[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