What is the solution for y^{'''}-3y^{''}-4y^'=0 ?

The solution for y^{'''}-3y^{''}-4y^'=0 is y=c_{1}+c_{2}e^{-t}+c_{3}e^{4t}

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y^{'''}-3y^{''}-4y^'=0

y^{^{'''}} - 3 y^{^{''}} - 4 y^{^{'}} = 0

y = c_{1} + c_{2} e^{- t} + c_{3} e^{4 t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{4 t}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{4 t}

f (t) = 106 - 80 e^{- 0.4 t}

\int 3^{x} d x

2 + 3 + cos^{2} (x)

\int \frac{2}{x ^{4}} d x

\int 4 y^{3} e^{y} d y

What is the solution for y^{'''}-3y^{''}-4y^'=0 ?
The solution for y^{'''}-3y^{''}-4y^'=0 is y=c_{1}+c_{2}e^{-t}+c_{3}e^{4t}