What is the solution for 18xy^'-ln(x^6)=0,y(1)=12 ?

The solution for 18xy^'-ln(x^6)=0,y(1)=12 is y= 1/6 ln^2(x)+12

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18xy^'-ln(x^6)=0,y(1)=12

18 x y^{^{'}} - ln (x^{6}) = 0, y (1) = 12

y = \frac{1}{6} ln^{2} (x) + 12

Solution steps

18 x y^{'} (x) - ln (x^{6}) = 0

Isolate

y^{'} (x) : y^{'} (x) = \frac{ln ( x )}{3 x}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{ln ( x )}{3 x} d x

\int \frac{ln ( x )}{3 x} d x = \frac{1}{6} ln^{2} (x) + c_{1}

y = \frac{1}{6} ln^{2} (x) + c_{1}

Apply initial conditions:

y = \frac{1}{6} ln^{2} (x) + 12

y = \frac{1}{6} ln^{2} (x) + 12

f (x) = x^{2} 3 x + 5

\frac{\partial}{\partial x} (- arctan (\frac{x}{y}))

n = 0 \sum \infty (\frac{5}{4})^{n}

(8.1) (2.6)

cos (a^{4} + x^{4})

What is the solution for 18xy^'-ln(x^6)=0,y(1)=12 ?
The solution for 18xy^'-ln(x^6)=0,y(1)=12 is y= 1/6 ln^2(x)+12