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

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

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

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

y = e^{3 t} (6 cos (4 t) - 5 sin (4 t))

Solution steps

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

Solve linear ODE:

y = e^{3 t} (6 cos (4 t) - 5 sin (4 t))

y = e^{3 t} (6 cos (4 t) - 5 sin (4 t))

y^{^{''}} - 5 y^{^{'}} - 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 3

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 5, y^{^{'}} (0) = 33

y^{^{''}} - 16 y = 0, y (0) = 3

2 y^{^{''}} - 5 y^{^{'}} - 3 y = 3 y

10000 y^{^{''}} - 81 y^{^{'}} = 0, y (0) = 8, y^{^{'}} (0) = 5

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