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

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

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

\frac{\partial}{\partial x} (2 y (y^{2} + e^{2 x}))

\int_{0}^{N} e^{(- s t)} sin (2 t) d t

h \to 3 lim (\frac{3 ( x + h ) ^{3} - 3 x ^{3}}{h})

\int_{- 1}^{2} \frac{2 x - 7}{x ^{3}} d x

\frac{1}{1 - x}, 3

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