What is the solution for (y^'-e^{-t}+5)/y =-5,y(0)=-4 ?

The solution for (y^'-e^{-t}+5)/y =-5,y(0)=-4 is y= 1/(4e^t)-1-(13)/(4e^{5t)}

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(y^'-e^{-t}+5)/y =-5,y(0)=-4

\frac{y ^{^{'}} - e ^{- t} + 5}{y} = - 5, y (0) = - 4

y = \frac{1}{4 e ^{t}} - 1 - \frac{13}{4 e ^{5 t}}

Solution steps

\frac{y ^{'} ( t ) - e ^{- t} + 5}{y} = - 5

Solve linear ODE:

y = \frac{1}{4 e ^{t}} - 1 - \frac{13}{4 e ^{5 t}}

y = \frac{1}{4 e ^{t}} - 1 - \frac{13}{4 e ^{5 t}}

(5, - 1), (- 3, 4)

\frac{d}{d x} (\frac{x - 3}{x ^{2} - 3 x + 2})

\int 4 y^{4} d y

x = y (2 - y), x = 0

\frac{d}{d x} ((6 - x)^{\frac{1}{3}})

What is the solution for (y^'-e^{-t}+5)/y =-5,y(0)=-4 ?
The solution for (y^'-e^{-t}+5)/y =-5,y(0)=-4 is y= 1/(4e^t)-1-(13)/(4e^{5t)}