What is the solution for y^{''}+2y^'+5y=1,y^'(0)=0,y(0)=0 ?

The solution for y^{''}+2y^'+5y=1,y^'(0)=0,y(0)=0 is y=e^{-t}(-1/5 cos(2t)-1/10 sin(2t))+1/5

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y^{''}+2y^'+5y=1,y^'(0)=0,y(0)=0

y^{^{''}} + 2 y^{^{'}} + 5 y = 1, y^{^{'}} (0) = 0, y (0) = 0

y = e^{- t} (- \frac{1}{5} cos (2 t) - \frac{1}{10} sin (2 t)) + \frac{1}{5}

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 1

Solve linear ODE:

y = e^{- t} (- \frac{1}{5} cos (2 t) - \frac{1}{10} sin (2 t)) + \frac{1}{5}

y = e^{- t} (- \frac{1}{5} cos (2 t) - \frac{1}{10} sin (2 t)) + \frac{1}{5}

y^{^{''}} - 2 y^{^{'}} + y = x^{3} + 4 x

y^{^{''''}} - y = e^{2 t}

- \frac{1}{9} y^{^{''}} + \frac{2}{3} y^{^{'}} - y = \frac{t}{t + 1}

y^{^{''}} - 2 y^{^{'}} + y = e^{- x}

y^{^{''}} + 6 y^{^{'}} + 9 y = t e^{- 3 t}

What is the solution for y^{''}+2y^'+5y=1,y^'(0)=0,y(0)=0 ?
The solution for y^{''}+2y^'+5y=1,y^'(0)=0,y(0)=0 is y=e^{-t}(-1/5 cos(2t)-1/10 sin(2t))+1/5