What is the solution for (y^'-e^{-t}+6)/y =-6 ?

The solution for (y^'-e^{-t}+6)/y =-6 is y= 1/(5e^t)-1+(c_{1})/(e^{6t)}

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(y^'-e^{-t}+6)/y =-6

\frac{y ^{^{'}} - e ^{- t} + 6}{y} = - 6

y = \frac{1}{5 e ^{t}} - 1 + \frac{c _{1}}{e ^{6 t}}

Solution steps

\frac{y ^{'} ( t ) - e ^{- t} + 6}{y} = - 6

Solve linear ODE:

y = \frac{1}{5 e ^{t}} - 1 + \frac{c _{1}}{e ^{6 t}}

y = \frac{1}{5 e ^{t}} - 1 + \frac{c _{1}}{e ^{6 t}}

2 t y^{^{'}} - y = 2 t y^{3} cos (t)

ln (x e^{x^{9}})

y = 3 x^{4}, (- 2, 48)

\frac{d}{d x} (t^{x})

\frac{d}{d x} (sec (\frac{π}{3}))

What is the solution for (y^'-e^{-t}+6)/y =-6 ?
The solution for (y^'-e^{-t}+6)/y =-6 is y= 1/(5e^t)-1+(c_{1})/(e^{6t)}