What is the solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=8 ?

The solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=8 is y=e^t(2cos(2t)+3sin(2t))

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y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=8

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 8

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

4 y^{^{'''}} + 4 y^{^{''}} + y^{^{'}} = 0

y^{^{'''}} - 2 y^{^{''}} - 7 y^{^{'}} - 4 y = 0

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

y^{^{''}} - y = 0, y (0) = 5, y^{^{'}} (0) = 1

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 1

What is the solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=8 ?
The solution for y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=8 is y=e^t(2cos(2t)+3sin(2t))