What is the solution for 4y^{''}-16y^'+18y=0 ?

The solution for 4y^{''}-16y^'+18y=0 is y=e^{2t}(c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2))))

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4y^{''}-16y^'+18y=0

4 y^{^{''}} - 16 y^{^{'}} + 18 y = 0

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

Solution steps

4 y^{''} (t) - 16 y^{'} (t) + 18 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (\frac{25 x ^{2} - 1}{x + 3})

\int_{0}^{1} (e^{- 0.1 x}) d x

\frac{d}{d x} ((x + 2)^{3})

z \to 0 lim (z^{3} e^{\frac{1}{z}})

\int_{1}^{2} \frac{e ^{\frac{3}{x}}}{x ^{2}} d x

What is the solution for 4y^{''}-16y^'+18y=0 ?
The solution for 4y^{''}-16y^'+18y=0 is y=e^{2t}(c_{1}cos(t/(sqrt(2)))+c_{2}sin(t/(sqrt(2))))