What is the solution for y^'=4y+y^4,y(0)=1 ?

The solution for y^'=4y+y^4,y(0)=1 is y=\sqrt[3]{(4e^{12t})/(5-e^{12t)}}

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y^'=4y+y^4,y(0)=1

y^{^{'}} = 4 y + y^{4}, y (0) = 1

y = 3 \frac{4 e ^{12 t}}{5 - e ^{12 t}}

Solution steps

y^{'} (t) = 4 y + y^{4}

Solve Bernoulli ODE:

y = 3 \frac{4 e ^{12 t}}{5 - e ^{12 t}}

y = 3 \frac{4 e ^{12 t}}{5 - e ^{12 t}}

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\frac{d}{d x} (4 x^{3} - 168 x^{2} + 1728 x)

x \to - 2 + lim (\frac{x - 1}{3 x + 6})

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What is the solution for y^'=4y+y^4,y(0)=1 ?
The solution for y^'=4y+y^4,y(0)=1 is y=\sqrt[3]{(4e^{12t})/(5-e^{12t)}}