What is the solution for (y^'-e^{-t}+6)/y =-6,y(0)=1 ?

The solution for (y^'-e^{-t}+6)/y =-6,y(0)=1 is y= 1/(5e^t)-1+9/(5e^{6t)}

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(y^'-e^{-t}+6)/y =-6,y(0)=1

\frac{y ^{^{'}} - e ^{- t} + 6}{y} = - 6, y (0) = 1

y = \frac{1}{5 e ^{t}} - 1 + \frac{9}{5 e ^{6 t}}

Solution steps

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

Solve linear ODE:

y = \frac{1}{5 e ^{t}} - 1 + \frac{9}{5 e ^{6 t}}

y = \frac{1}{5 e ^{t}} - 1 + \frac{9}{5 e ^{6 t}}

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y = ln ∣ sin (x) ∣

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\int \frac{81}{x ^{3} - 9 x ^{2}} d x

What is the solution for (y^'-e^{-t}+6)/y =-6,y(0)=1 ?
The solution for (y^'-e^{-t}+6)/y =-6,y(0)=1 is y= 1/(5e^t)-1+9/(5e^{6t)}