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

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

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

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

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

y^{^{''}} + 4 y^{1} - y = 0

y^{^{''}} + π^{2} \cdot y = 0

y^{^{''}} + 16 y^{^{'}} + 100 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + (\frac{25}{6}) y^{^{'}} + 25 y = 0

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