What is the solution for 10y^{''}+20y^'+6260y=100 ?

The solution for 10y^{''}+20y^'+6260y=100 is y=e^{-t}(c_{1}cos(25t)+c_{2}sin(25t))+5/313

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10y^{''}+20y^'+6260y=100

10 y^{^{''}} + 20 y^{^{'}} + 6260 y = 100

y = e^{- t} (c_{1} cos (25 t) + c_{2} sin (25 t)) + \frac{5}{313}

Solution steps

10 y^{''} (t) + 20 y^{'} (t) + 6260 y = 100

Solve linear ODE:

y = e^{- t} (c_{1} cos (25 t) + c_{2} sin (25 t)) + \frac{5}{313}

y = e^{- t} (c_{1} cos (25 t) + c_{2} sin (25 t)) + \frac{5}{313}

y^{^{''}} + 2 y^{^{'}} - 8 y = 24 e^{2 t}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{3} + x

y^{^{''}} + 10 y^{^{'}} + 16 = 0

y^{^{''}} + 153.06 y^{^{'}} + 12739.6 y = 191094.97

\frac{d ^{3} y}{d x ^{3}} + y = e^{- x} + 1

What is the solution for 10y^{''}+20y^'+6260y=100 ?
The solution for 10y^{''}+20y^'+6260y=100 is y=e^{-t}(c_{1}cos(25t)+c_{2}sin(25t))+5/313