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

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

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y^{''}+2y^'+26y=0,y(0)=4,y^'(0)=-2

y^{^{''}} + 2 y^{^{'}} + 26 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

y = e^{- t} (4 cos (5 t) + \frac{2}{5} sin (5 t))

Solution steps

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

Solve linear ODE:

y = e^{- t} (4 cos (5 t) + \frac{2}{5} sin (5 t))

y = e^{- t} (4 cos (5 t) + \frac{2}{5} sin (5 t))

\frac{d}{d x} (x \cdot ln (x))

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

\int (15 x^{4}) d x

\frac{d}{d x} (\frac{- 7 x ^{2}}{2 x ^{2} - 5})

\frac{d}{d x} (- cos (5 x))

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