What is the solution for (y^'-e^{-t}+4)/y =-4 ?

The solution for (y^'-e^{-t}+4)/y =-4 is y= 1/(3e^t)-1+(c_{1})/(e^{4t)}

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(y^'-e^{-t}+4)/y =-4

\frac{y ^{^{'}} - e ^{- t} + 4}{y} = - 4

y = \frac{1}{3 e ^{t}} - 1 + \frac{c _{1}}{e ^{4 t}}

Solution steps

\frac{y ^{'} ( t ) - e ^{- t} + 4}{y} = - 4

Solve linear ODE:

y = \frac{1}{3 e ^{t}} - 1 + \frac{c _{1}}{e ^{4 t}}

y = \frac{1}{3 e ^{t}} - 1 + \frac{c _{1}}{e ^{4 t}}

\int sec (x) tan (x) - 5 e^{x} d x

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

sin (x y) d x = 2 x d y + 2 y d x

\int \frac{x ^{2}}{x + 2} d x

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

What is the solution for (y^'-e^{-t}+4)/y =-4 ?
The solution for (y^'-e^{-t}+4)/y =-4 is y= 1/(3e^t)-1+(c_{1})/(e^{4t)}