What is the solution for y^{''}+10y^'+34y=0,y(0)=1,y^'(0)=2 ?

The solution for y^{''}+10y^'+34y=0,y(0)=1,y^'(0)=2 is y=e^{-5t}(cos(3t)+7/3 sin(3t))

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y^{''}+10y^'+34y=0,y(0)=1,y^'(0)=2

y^{^{''}} + 10 y^{^{'}} + 34 y = 0, y (0) = 1, y^{^{'}} (0) = 2

y = e^{- 5 t} (cos (3 t) + \frac{7}{3} sin (3 t))

Solution steps

y^{''} (t) + 10 y^{'} (t) + 34 y = 0

Solve linear ODE:

y = e^{- 5 t} (cos (3 t) + \frac{7}{3} sin (3 t))

y = e^{- 5 t} (cos (3 t) + \frac{7}{3} sin (3 t))

y^{^{''}} - y = 0, y (1) = 0, y^{^{'}} (1) = - 1

x^{^{''}} - 26 x^{^{'}} + 25 x = 0, x (0) = - 1, x^{^{'}} (0) = 2

D^{2} (D + 4)^{3} y = 0

y^{^{'''}} - 8 y^{^{''}} + 15 y^{^{'}} = 0

y^{^{'''}} + 15 y^{^{''}} + 58 y^{^{'}} + 44 y = 0

What is the solution for y^{''}+10y^'+34y=0,y(0)=1,y^'(0)=2 ?
The solution for y^{''}+10y^'+34y=0,y(0)=1,y^'(0)=2 is y=e^{-5t}(cos(3t)+7/3 sin(3t))