What is the solution for (dy)/(dt)=5y+y^4,y(0)=1 ?

The solution for (dy)/(dt)=5y+y^4,y(0)=1 is y=\sqrt[3]{(5e^{15t})/(6-e^{15t)}}

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(dy)/(dt)=5y+y^4,y(0)=1

\frac{d y}{d t} = 5 y + y^{4}, y (0) = 1

y = 3 \frac{5 e ^{15 t}}{6 - e ^{15 t}}

Solution steps

\frac{d y}{d t} = 5 y + y^{4}

Solve Bernoulli ODE:

y = 3 \frac{5 e ^{15 t}}{6 - e ^{15 t}}

y = 3 \frac{5 e ^{15 t}}{6 - e ^{15 t}}

1 - cos^{2} (17 x)

f (x) = 6 x^{2} - 20 x^{3}

\frac{d}{d x} (5 - ∣ x - 2 ∣)

8 - x^{2}, x^{2}, - 3, 3

\frac{d}{d x} (\frac{2 2}{3} x^{\frac{3}{2}})

What is the solution for (dy)/(dt)=5y+y^4,y(0)=1 ?
The solution for (dy)/(dt)=5y+y^4,y(0)=1 is y=\sqrt[3]{(5e^{15t})/(6-e^{15t)}}