What is the solution for y^'=(t+y)^2-1,y(2)=6 ?

The solution for y^'=(t+y)^2-1,y(2)=6 is y=-8/(8t-17)-t

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y^'=(t+y)^2-1,y(2)=6

y^{^{'}} = (t + y)^{2} - 1, y (2) = 6

y = - \frac{8}{8 t - 17} - t

Solution steps

y^{'} (t) = (t + y)^{2} - 1

Solve using substitution:

y = - \frac{8}{8 t - 17} - t

y = - \frac{8}{8 t - 17} - t

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What is the solution for y^'=(t+y)^2-1,y(2)=6 ?
The solution for y^'=(t+y)^2-1,y(2)=6 is y=-8/(8t-17)-t