What is the solution for (dy}{dx}=\frac{(1-3y)^2)/x ,y(1)=1 ?

The solution for (dy}{dx}=\frac{(1-3y)^2)/x ,y(1)=1 is y=-(2ln(x))/(-6ln(x)+1)+1/(-6ln(x)+1)

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(dy}{dx}=\frac{(1-3y)^2)/x ,y(1)=1

\frac{d y}{d x} = \frac{( 1 - 3 y ) ^{2}}{x}, y (1) = 1

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

Solution steps

\frac{d y}{d x} = \frac{( 1 - 3 y ) ^{2}}{x}

Solve separable ODE:

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

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

\frac{d}{d x} ((x^{2} + 1)^{2})

n = 0 \sum \infty \frac{n ^{2}}{n !}

\frac{d}{d x} (sin (3 x) ln (sech (2 x)))

\frac{\partial}{\partial y} (ln (x^{2} + y^{2} + 2))

n = 1 \sum \infty \frac{x ^{n}}{1 ^{n}}

What is the solution for (dy}{dx}=\frac{(1-3y)^2)/x ,y(1)=1 ?
The solution for (dy}{dx}=\frac{(1-3y)^2)/x ,y(1)=1 is y=-(2ln(x))/(-6ln(x)+1)+1/(-6ln(x)+1)