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

The solution for y^'=-(1+y)^4,y(1)=2 is y= 3/(\sqrt[3]{81t-80)}-1

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y^'=-(1+y)^4,y(1)=2

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

y = \frac{3}{3 81 t - 80} - 1

Solution steps

y^{'} (t) = - (1 + y)^{4}

Solve separable ODE:

y = \frac{3}{3 81 t - 80} - 1

y = \frac{3}{3 81 t - 80} - 1

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What is the solution for y^'=-(1+y)^4,y(1)=2 ?
The solution for y^'=-(1+y)^4,y(1)=2 is y= 3/(\sqrt[3]{81t-80)}-1