What is the solution for y^'=(t+y)^2-1,y(2)=4 ?

The solution for y^'=(t+y)^2-1,y(2)=4 is y=-6/(6t-13)-t

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y^'=(t+y)^2-1,y(2)=4

y^{^{'}} = (t + y)^{2} - 1, y (2) = 4

y = - \frac{6}{6 t - 13} - t

Solution steps

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

Solve using substitution:

y = - \frac{6}{6 t - 13} - t

y = - \frac{6}{6 t - 13} - t

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What is the solution for y^'=(t+y)^2-1,y(2)=4 ?
The solution for y^'=(t+y)^2-1,y(2)=4 is y=-6/(6t-13)-t