What is the solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-10 ?

The solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-10 is y=e^{-4t}(3cos(3t)+2/3 sin(3t))

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y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-10

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 10

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

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

x \to - 2 + lim (\frac{- 7 x}{x + 2})

\int 7 \cdot tan^{3} (x) sec (x) d x

f (x) = 4 x^{2} + 9 x

\int \frac{6 t - 5}{t + 1} d t

f (x) = 5 x^{3}

What is the solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-10 ?
The solution for y^{''}+8y^'+25y=0,y(0)=3,y^'(0)=-10 is y=e^{-4t}(3cos(3t)+2/3 sin(3t))