What is the solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=5 ?

The solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=5 is y=e^{-t}(3cos(2t)+4sin(2t))

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y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=5

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = 5

y = e^{- t} (3 cos (2 t) + 4 sin (2 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{- t} (3 cos (2 t) + 4 sin (2 t))

y = e^{- t} (3 cos (2 t) + 4 sin (2 t))

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

y^{^{''''}} + 8 y^{^{'''}} + 24 y^{^{''}} + 32 y^{^{'}} + 16 y = 0

\frac{d ^{2} y}{d x ^{2}} + 5 \frac{d y}{d x} + 10 y = 0

θ^{^{''}} (t) + \frac{4 k}{3} θ (t) = 0

y^{^{''}} + 144 y = 0, y (0) = 3, y^{^{'}} (0) = - 48

What is the solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=5 ?
The solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=5 is y=e^{-t}(3cos(2t)+4sin(2t))