What is the solution for y^{''}+2y^'+10y=10 ?

The solution for y^{''}+2y^'+10y=10 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))+1

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y^{''}+2y^'+10y=10

y^{^{''}} + 2 y^{^{'}} + 10 y = 10

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + 1

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 y = 10

Solve linear ODE:

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + 1

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + 1

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

y^{^{''}} + 25 y^{^{'}} = - 24 e^{- t}

y^{^{''}} + 4 y^{^{'}} + 4 y = sin (x), y (0) = 1, y^{^{'}} (0) = 0

(d^{2} - 2 d + 2) y = x + e^{x} cos (x)

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

What is the solution for y^{''}+2y^'+10y=10 ?
The solution for y^{''}+2y^'+10y=10 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))+1