What is the solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=4 ?

The solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=4 is y=e^{-t}(3cos(2t)+7/2 sin(2t))

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y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=4

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 3, y^{^{'}} (0) = 4

y = e^{- t} (3 cos (2 t) + \frac{7}{2} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (3 cos (2 t) + \frac{7}{2} sin (2 t))

y = e^{- t} (3 cos (2 t) + \frac{7}{2} sin (2 t))

y^{^{'''}} + y^{^{''}} - 4 y^{^{'}} + 6 y = 0

14 y^{^{''}} - y^{^{'}} - 3 y = 0

\frac{d ^{2} y}{d x ^{2}} (1) - 9 y = 0

4 y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + a y = 0, y^{^{'}} (π) = 0, y^{^{'}} (0) = 0

What is the solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=4 ?
The solution for y^{''}+2y^'+5y=0,y(0)=3,y^'(0)=4 is y=e^{-t}(3cos(2t)+7/2 sin(2t))