What is the solution for y^{''}+25y^'+100y=90 ?

The solution for y^{''}+25y^'+100y=90 is y=c_{1}e^{-5t}+c_{2}e^{-20t}+9/10

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y^{''}+25y^'+100y=90

y^{^{''}} + 25 y^{^{'}} + 100 y = 90

y = c_{1} e^{- 5 t} + c_{2} e^{- 20 t} + \frac{9}{10}

Solution steps

y^{''} (t) + 25 y^{'} (t) + 100 y = 90

Solve linear ODE:

y = c_{1} e^{- 5 t} + c_{2} e^{- 20 t} + \frac{9}{10}

y = c_{1} e^{- 5 t} + c_{2} e^{- 20 t} + \frac{9}{10}

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = 2 - 24 e^{x} + 40 e^{5 x}

y^{^{''}} - 16 y = e^{- 5 x} (- 8 - 9 x)

y^{^{'''}} - 3 y^{^{''}} + 2 y = e^{- 4 t}

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

y^{^{''}} + 25 y = 7 sec (5 x)

What is the solution for y^{''}+25y^'+100y=90 ?
The solution for y^{''}+25y^'+100y=90 is y=c_{1}e^{-5t}+c_{2}e^{-20t}+9/10