What is the solution for y^'=(t+y)^2-1,y(3)=6 ?

The solution for y^'=(t+y)^2-1,y(3)=6 is y=-9/(9t-28)-t

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y^'=(t+y)^2-1,y(3)=6

y^{^{'}} = (t + y)^{2} - 1, y (3) = 6

y = - \frac{9}{9 t - 28} - t

Solution steps

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

Solve using substitution:

y = - \frac{9}{9 t - 28} - t

y = - \frac{9}{9 t - 28} - t

\int \frac{( x ^{2} + 3 x - 4 )}{x ^{3} - 4 x ^{2} + 4 x} d x

y = 4 x^{2}, x = 1, y = 0

\frac{2 e ^{- 2 s}}{5 s ^{2} + 45}

y = e^{\frac{( 3 x ^{2} )}{2}}

\frac{x ^{63} - 1}{x - 1}

What is the solution for y^'=(t+y)^2-1,y(3)=6 ?
The solution for y^'=(t+y)^2-1,y(3)=6 is y=-9/(9t-28)-t