What is the solution for y^'=(t+y)^2-1,y(2)=5 ?

The solution for y^'=(t+y)^2-1,y(2)=5 is y=-7/(7t-15)-t

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y^'=(t+y)^2-1,y(2)=5

y^{^{'}} = (t + y)^{2} - 1, y (2) = 5

y = - \frac{7}{7 t - 15} - t

Solution steps

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

Solve using substitution:

y = - \frac{7}{7 t - 15} - t

y = - \frac{7}{7 t - 15} - t

f (x) = \frac{1}{5 - x}, at x = 0

y = x^{3} - 14 x

\frac{d}{d x} ((x^{3} - 6 x) (2 - 4 x^{3}))

\int x (b - x)^{2} d x

\int \frac{( ln ( x ) ) ^{\frac{3}{2}}}{x} d x

What is the solution for y^'=(t+y)^2-1,y(2)=5 ?
The solution for y^'=(t+y)^2-1,y(2)=5 is y=-7/(7t-15)-t