What is the solution for y^'=x^3(1-y),y(0)=3 ?

The solution for y^'=x^3(1-y),y(0)=3 is y=1+2/(e^{\frac{x^4){4}}}

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y^'=x^3(1-y),y(0)=3

y^{^{'}} = x^{3} (1 - y), y (0) = 3

y = 1 + \frac{2}{e ^{\frac{x ^{4}}{4}}}

Solution steps

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

Solve linear ODE:

y = 1 + \frac{2}{e ^{\frac{x ^{4}}{4}}}

y = 1 + \frac{2}{e ^{\frac{x ^{4}}{4}}}

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

e^{- 1}

t \to 0 lim (e^{- 7 t})

\int - \frac{y}{3 + y ^{2}} d y

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

What is the solution for y^'=x^3(1-y),y(0)=3 ?
The solution for y^'=x^3(1-y),y(0)=3 is y=1+2/(e^{\frac{x^4){4}}}