What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-5 ?

The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-5 is y=e^{-3t}(2cos(4t)+1/4 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-5

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

y = e^{- 3 t} (2 cos (4 t) + \frac{1}{4} sin (4 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{- 3 t} (2 cos (4 t) + \frac{1}{4} sin (4 t))

y = e^{- 3 t} (2 cos (4 t) + \frac{1}{4} sin (4 t))

\int_{0}^{7} k (7 - x)

t \to + 0 lim (18 (1.15)^{t})

x \to 7 + lim (\frac{9 x}{x - 7})

\int 2 x + 3 e^{x} d x

x^{2} y^{^{''}} - x y^{^{'}} + y = ln (x)

What is the solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-5 ?
The solution for y^{''}+6y^'+25y=0,y(0)=2,y^'(0)=-5 is y=e^{-3t}(2cos(4t)+1/4 sin(4t))