What is the solution for 0.125y^{''}+4y^'+28y=0,y(0)=0 ?

The solution for 0.125y^{''}+4y^'+28y=0,y(0)=0 is y=-c_{2}e^{4(sqrt(2)-4)t}+c_{2}e^{-4(4+sqrt(2))t}

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0.125y^{''}+4y^'+28y=0,y(0)=0

0.125 y^{^{''}} + 4 y^{^{'}} + 28 y = 0, y (0) = 0

y = - c_{2} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

Solution steps

0.125 y^{''} (t) + 4 y^{'} (t) + 28 y = 0

Solve linear ODE:

y = - c_{2} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

y = - c_{2} e^{4 (2 - 4) t} + c_{2} e^{- 4 (4 + 2) t}

\int 2 x (x^{2} + 9)^{6} d x

\frac{d}{d x} (x^{4} - 6 x^{2})

\int (3 x^{2} + 7 x + 1) e^{x} d x

y^{^{''}} + 9 y = t^{2} \cdot e^{3 t} + 6

What is the solution for 0.125y^{''}+4y^'+28y=0,y(0)=0 ?
The solution for 0.125y^{''}+4y^'+28y=0,y(0)=0 is y=-c_{2}e^{4(sqrt(2)-4)t}+c_{2}e^{-4(4+sqrt(2))t}