What is the solution for y^{''}+8y^'+25y=0,y(0)=-3,y^'(0)=1 ?

The solution for y^{''}+8y^'+25y=0,y(0)=-3,y^'(0)=1 is y=e^{-4t}(-3cos(3t)-11/3 sin(3t))

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y^{''}+8y^'+25y=0,y(0)=-3,y^'(0)=1

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = 1

y = e^{- 4 t} (- 3 cos (3 t) - \frac{11}{3} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{- 4 t} (- 3 cos (3 t) - \frac{11}{3} sin (3 t))

y = e^{- 4 t} (- 3 cos (3 t) - \frac{11}{3} sin (3 t))

What is the solution for y^{''}+8y^'+25y=0,y(0)=-3,y^'(0)=1 ?
The solution for y^{''}+8y^'+25y=0,y(0)=-3,y^'(0)=1 is y=e^{-4t}(-3cos(3t)-11/3 sin(3t))