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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 7 y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 4 y = 0

2 y^{^{''}} - y^{^{'}} - 3 y = 0, y (2) = 5, y^{^{'}} (2) = 0

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

y^{^{''}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 5

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