What is the solution for x^{''}+2x^'+26x=0,x(0)=1,x^'(0)=4 ?

The solution for x^{''}+2x^'+26x=0,x(0)=1,x^'(0)=4 is x=e^{-t}(cos(5t)+sin(5t))

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x^{''}+2x^'+26x=0,x(0)=1,x^'(0)=4

x^{^{''}} + 2 x^{^{'}} + 26 x = 0, x (0) = 1, x^{^{'}} (0) = 4

x = e^{- t} (cos (5 t) + sin (5 t))

Solution steps

x^{''} (t) + 2 x^{'} (t) + 26 x = 0

Solve linear ODE:

x = e^{- t} (cos (5 t) + sin (5 t))

x = e^{- t} (cos (5 t) + sin (5 t))

y^{^{''''}} + y^{^{'''}} - 7 y^{^{''}} - y^{^{'}} + 6 y = 0

\frac{d ^{2} y}{d x ^{2}} = - k^{2} y

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

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

2 y^{^{''}} - 8 y^{^{'}} + 6 y = 0

What is the solution for x^{''}+2x^'+26x=0,x(0)=1,x^'(0)=4 ?
The solution for x^{''}+2x^'+26x=0,x(0)=1,x^'(0)=4 is x=e^{-t}(cos(5t)+sin(5t))