What is the solution for y^{''}+2y^'+10y=25^2+3 ?

The solution for y^{''}+2y^'+10y=25^2+3 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))+314/5

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y^{''}+2y^'+10y=25^2+3

y^{^{''}} + 2 y^{^{'}} + 10 y = 2 5^{2} + 3

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{314}{5}

Solution steps

y^{''} (t) + 2 y^{'} (t) + 10 y = 2 5^{2} + 3

Solve linear ODE:

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{314}{5}

y = e^{- t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{314}{5}

y^{^{''}} - 2 y^{^{'}} + 2 y = 4 x + 5, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} + 1 y = (sec (x)) tan (x)

y^{^{''}} - y^{^{'}} - 2 y = 13 x + 9

x^{^{''}} + 7 x^{^{'}} + 6 x = sin (2 t)

y^{^{''}} - 25 y = e^{5 x}

What is the solution for y^{''}+2y^'+10y=25^2+3 ?
The solution for y^{''}+2y^'+10y=25^2+3 is y=e^{-t}(c_{1}cos(3t)+c_{2}sin(3t))+314/5