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

The solution for y^'=(t+y)^2-1,y(1)=6 is y=-7/(7t-8)-t

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

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

y = - \frac{7}{7 t - 8} - t

Solution steps

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

Solve using substitution:

y = - \frac{7}{7 t - 8} - t

y = - \frac{7}{7 t - 8} - t

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What is the solution for y^'=(t+y)^2-1,y(1)=6 ?
The solution for y^'=(t+y)^2-1,y(1)=6 is y=-7/(7t-8)-t