What is the solution for y^{''}+6y^'+34y=0,y(0)=4,y^'(0)=-11 ?

The solution for y^{''}+6y^'+34y=0,y(0)=4,y^'(0)=-11 is y=e^{-3t}(4cos(5t)+1/5 sin(5t))

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y^{''}+6y^'+34y=0,y(0)=4,y^'(0)=-11

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 4, y^{^{'}} (0) = - 11

y = e^{- 3 t} (4 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} (4 cos (5 t) + \frac{1}{5} sin (5 t))

y = e^{- 3 t} (4 cos (5 t) + \frac{1}{5} sin (5 t))

x^{3} y^{^{'}} + 3 x^{2} y = \frac{1}{x}

\frac{d}{d x} (5 x^{2} sin (x) + 4 x)

f (x) = \frac{x ^{4} + c _{1}}{4 x ^{2}}

What is the solution for y^{''}+6y^'+34y=0,y(0)=4,y^'(0)=-11 ?
The solution for y^{''}+6y^'+34y=0,y(0)=4,y^'(0)=-11 is y=e^{-3t}(4cos(5t)+1/5 sin(5t))