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

The solution for y^{''}+5y^'=e^{-t},y(0)=0,y^'(0)=1 is y= 3/20+1/4-3/20 e^{-5t}-1/4 e^{-t}

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y^{''}+5y^'=e^{-t},y(0)=0,y^'(0)=1

y^{^{''}} + 5 y^{^{'}} = e^{- t}, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{3}{20} + \frac{1}{4} - \frac{3}{20} e^{- 5 t} - \frac{1}{4} e^{- t}

Solution steps

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

Solve linear ODE:

y = \frac{3}{20} + \frac{1}{4} - \frac{3}{20} e^{- 5 t} - \frac{1}{4} e^{- t}

y = \frac{3}{20} + \frac{1}{4} - \frac{3}{20} e^{- 5 t} - \frac{1}{4} e^{- t}

\frac{\partial}{\partial y} (e^{x y} ln (z))

\int \frac{( ln ( 2 x ) )}{x} d x

x \to 4 lim (f (x))

f (x) = - \frac{1}{x}

\int \frac{1}{x ^{2} + 2 x + 1} d x

What is the solution for y^{''}+5y^'=e^{-t},y(0)=0,y^'(0)=1 ?
The solution for y^{''}+5y^'=e^{-t},y(0)=0,y^'(0)=1 is y= 3/20+1/4-3/20 e^{-5t}-1/4 e^{-t}