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

The solution for y^{''}-3y^'-4y=e^{-t},y(0)=0,y^'(0)=-1 is y=-4/25 e^{4t}+4/25 e^{-t}-(e^{-t}t)/5

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y^{''}-3y^'-4y=e^{-t},y(0)=0,y^'(0)=-1

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = - 1

y = - \frac{4}{25} e^{4 t} + \frac{4}{25} e^{- t} - \frac{e ^{- t} t}{5}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 4 y = e^{- t}

Solve linear ODE:

y = - \frac{4}{25} e^{4 t} + \frac{4}{25} e^{- t} - \frac{e ^{- t} t}{5}

y = - \frac{4}{25} e^{4 t} + \frac{4}{25} e^{- t} - \frac{e ^{- t} t}{5}

(d^{2} - 2 d + 5) y = e^{- x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 3 x^{2} e^{2 x} sin (2 x)

y^{^{''}} + 2 y + 5 y = 5 sin (2 t)

y^{^{''}} + y = - 8 e^{3 t}

y^{^{''}} - 3 y^{^{'}} - 4 y = - 8 e^{t} (cos (2 t))

What is the solution for y^{''}-3y^'-4y=e^{-t},y(0)=0,y^'(0)=-1 ?
The solution for y^{''}-3y^'-4y=e^{-t},y(0)=0,y^'(0)=-1 is y=-4/25 e^{4t}+4/25 e^{-t}-(e^{-t}t)/5