What is the solution for y^'=5y+y^5,y(0)=1 ?

The solution for y^'=5y+y^5,y(0)=1 is y=\sqrt[4]{(5e^{20t})/(6-e^{20t)}}

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y^'=5y+y^5,y(0)=1

y^{^{'}} = 5 y + y^{5}, y (0) = 1

y = 4 \frac{5 e ^{20 t}}{6 - e ^{20 t}}

Solution steps

y^{'} (t) = 5 y + y^{5}

Solve Bernoulli ODE:

y = 4 \frac{5 e ^{20 t}}{6 - e ^{20 t}}

y = 4 \frac{5 e ^{20 t}}{6 - e ^{20 t}}

x \to 2 - lim (x^{2} + 3 x)

\int x - \frac{1}{2} x + \frac{2}{x} d x

\int sec (\frac{1}{2}) x tan (\frac{1}{2}) x d x

\frac{\partial}{\partial x} (2 (\frac{y}{x})^{\frac{1}{2}})

f (x) = 3 x + 3

What is the solution for y^'=5y+y^5,y(0)=1 ?
The solution for y^'=5y+y^5,y(0)=1 is y=\sqrt[4]{(5e^{20t})/(6-e^{20t)}}