What is the solution for y^'=(6x^4+y^4)/(xy^3) ?

The solution for y^'=(6x^4+y^4)/(xy^3) is y=x\sqrt[4]{24ln(x)+c_{1}},y=-x\sqrt[4]{24ln(x)+c_{1}}

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y^'=(6x^4+y^4)/(xy^3)

y^{^{'}} = \frac{6 x ^{4} + y ^{4}}{x y ^{3}}

Solution steps

y^{'} (x) = \frac{6 x ^{4} + y ^{4}}{x y ^{3}}

Solve Bernoulli ODE:

\int \frac{7 x ^{2}}{5 - x ^{6}} d x

x = - 3, x = 1, y = e^{x}, y = 1 - e^{x}

x y^{^{'}} + y = e^{x}, y (1) = 3

x \to 0 lim (x^{2} sin (x^{2}))

\frac{d}{d x} (x^{3} cos (3 x))

What is the solution for y^'=(6x^4+y^4)/(xy^3) ?
The solution for y^'=(6x^4+y^4)/(xy^3) is y=x\sqrt[4]{24ln(x)+c_{1}},y=-x\sqrt[4]{24ln(x)+c_{1}}