What is the solution for y^{''}-4y^'+8y=0,y(0)=0,y^'(0)=2 ?

The solution for y^{''}-4y^'+8y=0,y(0)=0,y^'(0)=2 is y=e^{2t}sin(2t)

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y^{''}-4y^'+8y=0,y(0)=0,y^'(0)=2

y^{^{''}} - 4 y^{^{'}} + 8 y = 0, y (0) = 0, y^{^{'}} (0) = 2

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = 0, y^{^{'}} (0) = 6, y (1) = 8

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

y^{^{''}} + 4 y^{^{'}} + 6.25 y = 0

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

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 0, y^{^{'}} (0) = - 9

What is the solution for y^{''}-4y^'+8y=0,y(0)=0,y^'(0)=2 ?
The solution for y^{''}-4y^'+8y=0,y(0)=0,y^'(0)=2 is y=e^{2t}sin(2t)