What is the solution for 4xy^3y^'=x^4+4y^4 ?

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

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4xy^3y^'=x^4+4y^4

4 x y^{3} y^{^{'}} = x^{4} + 4 y^{4}

Solution steps

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

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