What is the solution for linear y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=4 ?

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

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linear y^{''}-2y^'+5y=0,y(0)=2,y^'(0)=4

linear

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = 2, y^{^{'}} (0) = 4

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

Solution steps

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

Solve linear ODE:

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

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

\frac{1}{4} x^{^{''}} + 2 x^{^{'}} + 4 x = 0

8 y^{^{''}} + y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

y^{^{''}} + 5 y^{^{'}} - 8 y = 0

y^{^{''}} + 20 y^{^{'}} + 64 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{1}{8} x^{^{''}} + 36 x = 0

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