What is the solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-14 ?

The solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-14 is y=e^{-4t}(4cos(3t)+2/3 sin(3t))

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y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-14

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 14

y = e^{- 4 t} (4 cos (3 t) + \frac{2}{3} sin (3 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 4 t} (4 cos (3 t) + \frac{2}{3} sin (3 t))

y = e^{- 4 t} (4 cos (3 t) + \frac{2}{3} sin (3 t))

\int \frac{a}{x} d x

(a (t) sin^{3} (t))^{^{'}}

y = sin (7 x)

y^{^{''}} + 4 y^{^{'}} + \frac{17}{4} y = 0

\frac{\partial}{\partial x} (1 - cos (x))

What is the solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-14 ?
The solution for y^{''}+8y^'+25y=0,y(0)=4,y^'(0)=-14 is y=e^{-4t}(4cos(3t)+2/3 sin(3t))