What is the solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 ?

The solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 is y=-7/2 e^{-t}sin(2t)

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y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = - 7

y = - \frac{7}{2} e^{- t} sin (2 t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = - \frac{7}{2} e^{- t} sin (2 t)

y = - \frac{7}{2} e^{- t} sin (2 t)

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 3, y^{^{'}} (0) = 4

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 9, y^{^{'}} (0) = 2

y^{^{''}} + 3 y^{^{'}} - 9 y = 0

100 y^{^{''}} + 90 y^{^{'}} - 162 y = 0

y^{^{''}} - 4 y^{^{'}} - 13 y = 0

What is the solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 ?
The solution for y^{''}+2y^'+5y=0,y(0)=0,y^'(0)=-7 is y=-7/2 e^{-t}sin(2t)