What is the solution for y^{''}+8y^'+25y=0,y(0)=0,y^'(0)=0.9 ?

The solution for y^{''}+8y^'+25y=0,y(0)=0,y^'(0)=0.9 is y=(0.9)/3 e^{-4t}sin(3t)

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y^{''}+8y^'+25y=0,y(0)=0,y^'(0)=0.9

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 0.9

y = \frac{0.9}{3} e^{- 4 t} sin (3 t)

Solution steps

y^{''} (t) + 8 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = \frac{0.9}{3} e^{- 4 t} sin (3 t)

y = \frac{0.9}{3} e^{- 4 t} sin (3 t)

y^{^{''}} - 8 y^{^{'}} + 25 y = 0, y (0) = e^{4 x} cos (3 x)

81 y^{^{''''}} = 256 y

\frac{d ^{2} y}{d x ^{2}} = 16 y

y^{^{''}} + 12 y^{^{'}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{'''}} - 13 y^{^{''}} + 36 y^{^{'}} = 0

What is the solution for y^{''}+8y^'+25y=0,y(0)=0,y^'(0)=0.9 ?
The solution for y^{''}+8y^'+25y=0,y(0)=0,y^'(0)=0.9 is y=(0.9)/3 e^{-4t}sin(3t)