What is the solution for y^{''}+6y^'+25y=0,y(0)=6,y^'(0)=-17 ?

The solution for y^{''}+6y^'+25y=0,y(0)=6,y^'(0)=-17 is y=e^{-3t}(6cos(4t)+1/4 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=6,y^'(0)=-17

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 6, y^{^{'}} (0) = - 17

y = e^{- 3 t} (6 cos (4 t) + \frac{1}{4} sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{- 3 t} (6 cos (4 t) + \frac{1}{4} sin (4 t))

y = e^{- 3 t} (6 cos (4 t) + \frac{1}{4} sin (4 t))

x^{^{''}} + 36 x = 0

y^{^{''}} + 4 y^{^{'}} + 2 y = 0, y^{^{'}} (0) = 1, y (0) = 0

\frac{d ^{2} y}{d x ^{2}} - 13 \frac{d y}{d x} + 12 y = 0

- 2 y^{^{''}} - 2 \cdot π^{2} y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = 4, y^{^{'}} (0) = 1

What is the solution for y^{''}+6y^'+25y=0,y(0)=6,y^'(0)=-17 ?
The solution for y^{''}+6y^'+25y=0,y(0)=6,y^'(0)=-17 is y=e^{-3t}(6cos(4t)+1/4 sin(4t))