What is the solution for y^{''}-4y^'-5y=125,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-4y^'-5y=125,y(0)=0,y^'(0)=0 is y=(-125/6+25)e^{5t}+125/6 e^{-t}-25

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y^{''}-4y^'-5y=125,y(0)=0,y^'(0)=0

y^{^{''}} - 4 y^{^{'}} - 5 y = 125, y (0) = 0, y^{^{'}} (0) = 0

y = (- \frac{125}{6} + 25) e^{5 t} + \frac{125}{6} e^{- t} - 25

Solution steps

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

Solve linear ODE:

y = (- \frac{125}{6} + 25) e^{5 t} + \frac{125}{6} e^{- t} - 25

y = (- \frac{125}{6} + 25) e^{5 t} + \frac{125}{6} e^{- t} - 25

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

y^{^{''}} - y = \frac{1}{e ^{x} + 1}

(d^{4} + 4 d^{2}) y = x^{2} + 1

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

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} (x^{2} - 4 x + 5)

What is the solution for y^{''}-4y^'-5y=125,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'-5y=125,y(0)=0,y^'(0)=0 is y=(-125/6+25)e^{5t}+125/6 e^{-t}-25