What is the solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-5 ?

The solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-5 is y=e^{-3t}(2cos(5t)+1/5 sin(5t))

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y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-5

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

y = e^{- 3 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 34 y = 0

Solve linear ODE:

y = e^{- 3 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

y = e^{- 3 t} (2 cos (5 t) + \frac{1}{5} sin (5 t))

9 y^{^{''}} + 16 y = 0

a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = 0

2 y^{^{''}} + 16 y = 0

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 6 x = 0

y^{^{''}} - 15 y^{^{'}} + 54 y = 0

What is the solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-5 ?
The solution for y^{''}+6y^'+34y=0,y(0)=2,y^'(0)=-5 is y=e^{-3t}(2cos(5t)+1/5 sin(5t))