What is the solution for f^{''}(t)+2f^'(t)+5f(t)=0,f(0)=3,f^'(0)=-7 ?

The solution for f^{''}(t)+2f^'(t)+5f(t)=0,f(0)=3,f^'(0)=-7 is f(t)=e^{-t}(3cos(2t)-2sin(2t))

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f^{''}(t)+2f^'(t)+5f(t)=0,f(0)=3,f^'(0)=-7

f^{^{''}} (t) + 2 f^{^{'}} (t) + 5 f (t) = 0, f (0) = 3, f^{^{'}} (0) = - 7

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

\frac{d ^{2} x}{d t ^{2}} - 6 \frac{d x}{d t} + 9 x = 0

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

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y^{^{'}} (- 1) = 9, y (- 1) = 3

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