What is the solution for 2y^{''}+12y^'+50y=0,y(0)=0,y^'(0)=-8 ?

The solution for 2y^{''}+12y^'+50y=0,y(0)=0,y^'(0)=-8 is y=-2e^{-3t}sin(4t)

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2y^{''}+12y^'+50y=0,y(0)=0,y^'(0)=-8

2 y^{^{''}} + 12 y^{^{'}} + 50 y = 0, y (0) = 0, y^{^{'}} (0) = - 8

y = - 2 e^{- 3 t} sin (4 t)

Solution steps

2 y^{''} (t) + 12 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = - 2 e^{- 3 t} sin (4 t)

y = - 2 e^{- 3 t} sin (4 t)

y = x + 8, (8, 4)

\int 8 e^{t} - 5 d t

\frac{8 x ^{2} + 18 x}{( 9 + 8 x ) ^{2}}

x^{\frac{1}{6}}

\frac{d}{d x} (- x^{2} + 13 x - 22)

What is the solution for 2y^{''}+12y^'+50y=0,y(0)=0,y^'(0)=-8 ?
The solution for 2y^{''}+12y^'+50y=0,y(0)=0,y^'(0)=-8 is y=-2e^{-3t}sin(4t)