What is the solution for -3y^'+(2y)/x =((y^4))/(x^4) ?

The solution for -3y^'+(2y)/x =((y^4))/(x^4) is y= x/(\sqrt[3]{-1+c_{1)x}}

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-3y^'+(2y)/x =((y^4))/(x^4)

- 3 y^{^{'}} + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

Solution steps

- 3 y^{'} (x) + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

Solve Bernoulli ODE:

\frac{d y}{d x} = \frac{x - y - 3}{x - 3}

\frac{1}{0.012 s ^{2} + 1.55 s + 50}

s \to 1 lim (\frac{( 2 s + 3 )}{s})

\frac{d y}{d x} = 2 y + 3

\int_{1}^{8} (x^{2} - 4 x + 6) d x

What is the solution for -3y^'+(2y)/x =((y^4))/(x^4) ?
The solution for -3y^'+(2y)/x =((y^4))/(x^4) is y= x/(\sqrt[3]{-1+c_{1)x}}