What is the solution for 3y^'+y=(1-2x)y^4 ?

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

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3y^'+y=(1-2x)y^4

3 y^{^{'}} + y = (1 - 2 x) y^{4}

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

Solution steps

3 y^{'} (x) + y = (1 - 2 x) y^{4}

Solve Bernoulli ODE:

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

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

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

\frac{d}{d x} (3 x - 4 x^{3} + 2^{x} - e + 7)

y = 1 + x^{3}, (2, 3)

\int \frac{1}{( x - 2 ) ln ( x - 2 )} d x

\int (16 x^{3} - 3 x^{2} + 2) d x

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