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

The solution for y^{''}-4y^'+8y=0,y^'(0)=2 is y=e^{2t}((1-c_{2})cos(2t)+c_{2}sin(2t))

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

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

y = e^{2 t} ((1 - c_{2}) cos (2 t) + c_{2} sin (2 t))

Solution steps

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

Solve linear ODE:

y = e^{2 t} ((1 - c_{2}) cos (2 t) + c_{2} sin (2 t))

y = e^{2 t} ((1 - c_{2}) cos (2 t) + c_{2} sin (2 t))

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 16

y^{^{''}} + 2 y^{^{'}} + 10 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

6 x^{^{''}} + 8 x^{^{'}} + 4 x = 0

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (0) = 3, y^{^{'}} (0) = - 2

What is the solution for y^{''}-4y^'+8y=0,y^'(0)=2 ?
The solution for y^{''}-4y^'+8y=0,y^'(0)=2 is y=e^{2t}((1-c_{2})cos(2t)+c_{2}sin(2t))