What is the solution for y^'=e^{2t+y-1}-2 ?

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

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y^'=e^{2t+y-1}-2

y^{^{'}} = e^{2 t + y - 1} - 2

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

Solution steps

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

Solve using substitution:

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

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

\frac{\partial}{\partial x} (e^{x y} + ln (\frac{x}{y}))

\frac{\partial}{\partial x} (2 x y^{2} + 4)

w \to \infty lim (\frac{2 w ^{2} - 3 w + 4}{5 w ^{2} + 7 w - 1})

y^{^{''}} - k y = 0

x \to - 2 lim ((x - 6)^{\frac{2}{3}})

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