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

The solution for 2y^{''}-4y^'+8y=0 is y=e^t(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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

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

y = e^{t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y^{^{''}} + 3 y^{^{'}} + 2 y = t

f (x) = 3 (x)^{2} - 2 (x)

\frac{d y}{d x} = 4 - \frac{5 y}{300 - 3 x}

\int \frac{1}{x ^{2}} d x

x \to 1 lim (5 - 2 x)

What is the solution for 2y^{''}-4y^'+8y=0 ?
The solution for 2y^{''}-4y^'+8y=0 is y=e^t(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))