What is the solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=1 ?

The solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=1 is y=e^{-2t}(cos(t)+3sin(t))

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y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=1

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 4 y = 0, y (0) = - 5, y^{^{'}} (0) = 2

y^{^{''}} + y = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3

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

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 8 x = 0

y^{^{''''''}} + 3 y^{^{''''}} + 3 y^{^{''}} + y = 0

What is the solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=1 ?
The solution for y^{''}+4y^'+5y=0,y(0)=1,y^'(0)=1 is y=e^{-2t}(cos(t)+3sin(t))