What is the solution for y^{''}-12y^'+40y=0,y(0)=6,y^'(0)=-2 ?

The solution for y^{''}-12y^'+40y=0,y(0)=6,y^'(0)=-2 is y=e^{6t}(6cos(2t)-19sin(2t))

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y^{''}-12y^'+40y=0,y(0)=6,y^'(0)=-2

y^{^{''}} - 12 y^{^{'}} + 40 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

y = e^{6 t} (6 cos (2 t) - 19 sin (2 t))

Solution steps

y^{''} (t) - 12 y^{'} (t) + 40 y = 0

Solve linear ODE:

y = e^{6 t} (6 cos (2 t) - 19 sin (2 t))

y = e^{6 t} (6 cos (2 t) - 19 sin (2 t))

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

y^{^{''''}} + 12 y^{^{'''}} + 36 y^{^{''}} = 0

3 y^{^{''}} + 12 y^{^{'}} - 2 y = 0

y^{^{''}} + 200 y^{^{'}} + 1000 y = 0

y^{^{''}} - 4^{2} y = 0

What is the solution for y^{''}-12y^'+40y=0,y(0)=6,y^'(0)=-2 ?
The solution for y^{''}-12y^'+40y=0,y(0)=6,y^'(0)=-2 is y=e^{6t}(6cos(2t)-19sin(2t))