What is the solution for y^{''}+6y^'+25y=0,y(0)=0,y^'(0)=2 ?

The solution for y^{''}+6y^'+25y=0,y(0)=0,y^'(0)=2 is y= 1/2 e^{-3t}sin(4t)

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y^{''}+6y^'+25y=0,y(0)=0,y^'(0)=2

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y = \frac{1}{2} e^{- 3 t} sin (4 t)

Solution steps

y^{''} (t) + 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = \frac{1}{2} e^{- 3 t} sin (4 t)

y = \frac{1}{2} e^{- 3 t} sin (4 t)

4 \frac{d ^{2} y}{d x ^{2}} + 36 y = 0

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{\frac{2}{3} x}

y^{^{'''}} - 6 y^{^{''}} + y^{^{'}} - 6 y = 0

2 y - 3 y^{^{''}} = 0

y^{^{''}} + 4 y^{^{'}} - 7 y = 0

What is the solution for y^{''}+6y^'+25y=0,y(0)=0,y^'(0)=2 ?
The solution for y^{''}+6y^'+25y=0,y(0)=0,y^'(0)=2 is y= 1/2 e^{-3t}sin(4t)