What is the solution for y^{''}+4y^'+13y=0,y(0)=7,y^'(0)=7 ?

The solution for y^{''}+4y^'+13y=0,y(0)=7,y^'(0)=7 is y=e^{-2t}(7cos(3t)+7sin(3t))

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y^{''}+4y^'+13y=0,y(0)=7,y^'(0)=7

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 7, y^{^{'}} (0) = 7

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

f (x) = (3 x - 4)

\frac{d}{d x} (50 x)

ln (1 + x^{2})

f (x) = e^{2} + ln (7)

(\frac{1 - cos ( x )}{x})^{^{'}}

What is the solution for y^{''}+4y^'+13y=0,y(0)=7,y^'(0)=7 ?
The solution for y^{''}+4y^'+13y=0,y(0)=7,y^'(0)=7 is y=e^{-2t}(7cos(3t)+7sin(3t))