What is the solution for y^{''}-4y^'-5y=125 ?

The solution for y^{''}-4y^'-5y=125 is y=c_{1}e^{5t}+c_{2}e^{-t}-25

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y^{''}-4y^'-5y=125

y^{^{''}} - 4 y^{^{'}} - 5 y = 125

y = c_{1} e^{5 t} + c_{2} e^{- t} - 25

Solution steps

y^{''} (t) - 4 y^{'} (t) - 5 y = 125

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- t} - 25

y = c_{1} e^{5 t} + c_{2} e^{- t} - 25

y^{^{''}} + 4 y^{^{'}} + 5 y = - 5 x + 3 e^{- x}

y^{^{''}} + 4 y = 4

y^{^{''}} + y^{^{'}} - 12 y = 2 cos (t), y (0) = 1, y^{^{'}} (0) = 1

y^{^{'''}} + 8 y = 2 x - 5 + 8 e^{- 2 x}

y^{^{''}} - y = cosh (t)

What is the solution for y^{''}-4y^'-5y=125 ?
The solution for y^{''}-4y^'-5y=125 is y=c_{1}e^{5t}+c_{2}e^{-t}-25