What is the solution for -y+10e^{-t}=y^' ?

The solution for -y+10e^{-t}=y^' is y=(10t)/(e^t)+(c_{1})/(e^t)

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-y+10e^{-t}=y^'

- y + 10 e^{- t} = y^{^{'}}

y = \frac{10 t}{e ^{t}} + \frac{c _{1}}{e ^{t}}

Solution steps

- y + 10 e^{- t} = y^{'} (t)

Solve linear ODE:

y = \frac{10 t}{e ^{t}} + \frac{c _{1}}{e ^{t}}

y = \frac{10 t}{e ^{t}} + \frac{c _{1}}{e ^{t}}

\int ln (6 x) d x

sin (2 t) + cos (2 t)

\int sin^{3} (3 θ) cos (3 θ) d θ

x y^{^{'}} + y = - 2 x y^{2}, y (1) = - 3

\frac{d}{d x} (\frac{4}{11 - x})

What is the solution for -y+10e^{-t}=y^' ?
The solution for -y+10e^{-t}=y^' is y=(10t)/(e^t)+(c_{1})/(e^t)