What is the solution for y^{''}+6y^'+25y=0,y(0)=-3,y^'(0)=7 ?

The solution for y^{''}+6y^'+25y=0,y(0)=-3,y^'(0)=7 is y=e^{-3t}(-3cos(4t)-1/2 sin(4t))

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y^{''}+6y^'+25y=0,y(0)=-3,y^'(0)=7

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = 7

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} = 0

2 y^{^{''}} + 2 y^{^{'}} + 5 y = 0

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

x^{^{''}} + 8 x^{^{'}} + 4 x = 0, x (0) = 0, x^{^{'}} (0) = 1

x^{^{''}} - 4 a^{2} x = 0

What is the solution for y^{''}+6y^'+25y=0,y(0)=-3,y^'(0)=7 ?
The solution for y^{''}+6y^'+25y=0,y(0)=-3,y^'(0)=7 is y=e^{-3t}(-3cos(4t)-1/2 sin(4t))