What is the solution for y^'=(t+y)^2-1,y(3)=4 ?

The solution for y^'=(t+y)^2-1,y(3)=4 is y=-7/(7t-22)-t

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y^'=(t+y)^2-1,y(3)=4

y^{^{'}} = (t + y)^{2} - 1, y (3) = 4

y = - \frac{7}{7 t - 22} - t

Solution steps

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

Solve using substitution:

y = - \frac{7}{7 t - 22} - t

y = - \frac{7}{7 t - 22} - t

\int z (z^{2} - \frac{1}{4 z}) d z

\int_{0}^{1} x e^{x^{2}} d x

\int \frac{ln ^{5} ( x )}{x} d x

y^{^{'}} = \frac{( 2 y + 3 ) ^{2}}{( 4 x + 5 ) ^{2}}

\frac{\partial}{\partial y} ((x + 2 y)^{2})

What is the solution for y^'=(t+y)^2-1,y(3)=4 ?
The solution for y^'=(t+y)^2-1,y(3)=4 is y=-7/(7t-22)-t