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

The solution for -3y^'+(3y)/x =x^3y^4 is y=(\sqrt[3]{7}x)/(\sqrt[3]{x^7+c_{1)}}

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

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

Solution steps

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

Solve Bernoulli ODE:

\frac{d y}{d t} = 2 y + y^{3}, y (0) = 1

x^{\frac{2}{7}}

4 x - 3

y = \frac{4 x}{1 + x ^{2}}, (4, \frac{16}{17})

\int_{0}^{\frac{π}{2}} - 3 sin^{2} (3 x) d x

What is the solution for -3y^'+(3y)/x =x^3y^4 ?
The solution for -3y^'+(3y)/x =x^3y^4 is y=(\sqrt[3]{7}x)/(\sqrt[3]{x^7+c_{1)}}