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

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

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y^'+xy=xy^4

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

y = 3 \frac{1}{e ^{\frac{3 x ^{2}}{2} - c_{1}} + 1}

Solution steps

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

Solve Bernoulli ODE:

y = 3 \frac{1}{e ^{\frac{3 x ^{2}}{2} - c_{1}} + 1}

y = 3 \frac{1}{e ^{\frac{3 x ^{2}}{2} - c_{1}} + 1}

x \to 2 lim (- f (x))

\int_{0}^{2} x^{5} d x

\int cos^{4} (2 t) d t

f (x) = x^{4} + 4 x

\int (1 + tan (x))^{2} d x

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