What is the solution for y^'=(y+2t)^2 ?

The solution for y^'=(y+2t)^2 is y=tan(sqrt(2)(t+c_{1}))sqrt(2)-2t

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y^'=(y+2t)^2

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

y = tan (2 (t + c_{1})) 2 - 2 t

Solution steps

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

Solve using substitution:

y = tan (2 (t + c_{1})) 2 - 2 t

y = tan (2 (t + c_{1})) 2 - 2 t

- 2 y - y^{2} d x + (3 + 2 x - x^{2}) d y = 0

y^{^{''}} - y^{^{'}} - 2 y = cos (x) - sin (2 x)

\frac{d}{d x} (- 2 a sin (2 x) + 2 b cos (2 x))

y = 2 + 8 e^{6 x}

\frac{d}{d x} (\frac{( e ^{x} - 2 )}{( 1 - e ^{- x} ) ^{2}})

What is the solution for y^'=(y+2t)^2 ?
The solution for y^'=(y+2t)^2 is y=tan(sqrt(2)(t+c_{1}))sqrt(2)-2t