What is the solution for (dy)/(dt)=y-4t+(y-4t)^2+4 ?

The solution for (dy)/(dt)=y-4t+(y-4t)^2+4 is y=(e^{t+c_{1}})/(1-e^{t+c_{1)}}+4t

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(dy)/(dt)=y-4t+(y-4t)^2+4

\frac{d y}{d t} = y - 4 t + (y - 4 t)^{2} + 4

y = \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}} + 4 t

Solution steps

\frac{d y}{d t} = y - 4 t + (y - 4 t)^{2} + 4

Solve using substitution:

y = \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}} + 4 t

y = \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}} + 4 t

\int x (2 x^{2} + 1)^{5} d x

(- 3, 5), (2, - 1)

x \to - 2 - lim (\frac{3}{x ^{2} - 4})

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

a e^{- b^{2} x^{2}}

What is the solution for (dy)/(dt)=y-4t+(y-4t)^2+4 ?
The solution for (dy)/(dt)=y-4t+(y-4t)^2+4 is y=(e^{t+c_{1}})/(1-e^{t+c_{1)}}+4t