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

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

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

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

y = e^{3 t} (2 cos (4 t) + c_{2} sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{3 t} (2 cos (4 t) + c_{2} sin (4 t))

y = e^{3 t} (2 cos (4 t) + c_{2} sin (4 t))

(D^{3} + 6 D^{2} + 12 D + 8) y = 0

576 y^{^{''}} - 528 y^{^{'}} + 121 y = 0, y (0) = 9, y^{^{'}} (0) = 3

y^{^{''}} - 6 y^{^{'}} + 12 y^{^{'}} - 8 y = 0

x^{^{''}} + 16 x = 0, x (0) = - 1, x^{^{'}} (0) = - 6

4 y^{^{''}} + 20 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 2

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