What is the solution for y^'=((t^4+y^4))/(ty^3) ?

The solution for y^'=((t^4+y^4))/(ty^3) is y=t\sqrt[4]{4ln(t)+c_{1}},y=-t\sqrt[4]{4ln(t)+c_{1}}

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y^'=((t^4+y^4))/(ty^3)

y^{^{'}} = \frac{( t ^{4} + y ^{4} )}{t y ^{3}}

Solution steps

y^{'} (t) = \frac{( t ^{4} + y ^{4} )}{t y ^{3}}

Solve Bernoulli ODE:

\int \frac{3 - x}{x ^{2} - 1} d x

x \to - 1 - lim (\frac{x - 2}{x ^{2} - 1})

x \to + 0 lim (x)

e^{2 x}

\int (2 x + 3)^{4} d x

What is the solution for y^'=((t^4+y^4))/(ty^3) ?
The solution for y^'=((t^4+y^4))/(ty^3) is y=t\sqrt[4]{4ln(t)+c_{1}},y=-t\sqrt[4]{4ln(t)+c_{1}}