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

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

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

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{d ^{2} y}{d x ^{2}} + 81 y = 0

y^{^{''}} + 7 y^{^{'}} = 0

16 y^{^{''}} - 8 y^{^{'}} + 145 y = 0, y (0) = - 2, y^{^{'}} (0) = 1

(D^{5} - 2 D^{3} - 2 D^{2} - 3 D - 2) y = 0

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

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