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

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

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

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

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

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

f (x) = arcsin (\frac{x}{4})

\int x^{- \frac{1}{6}} d x

\int_{- 2}^{2} \frac{1}{4 - x ^{2}} d x

y = 6 x + \frac{10}{3}

\frac{d}{d x} (2 x^{\frac{1}{2}} y^{\frac{1}{2}})

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