What is the solution for 4y^{''}-16y^'+19y=0 ?

The solution for 4y^{''}-16y^'+19y=0 is y=e^{2t}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

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4y^{''}-16y^'+19y=0

4 y^{^{''}} - 16 y^{^{'}} + 19 y = 0

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

Solution steps

4 y^{''} (t) - 16 y^{'} (t) + 19 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (2 x^{2} cos (x))

\frac{y ^{^{'}}}{y} = x

\frac{6 x}{2 + x ^{2}}

\frac{\partial}{\partial y} (e^{2 y} sin (- 2 x))

f (x) = (2 x + 3)^{11} (4 - 5 x)^{9}

What is the solution for 4y^{''}-16y^'+19y=0 ?
The solution for 4y^{''}-16y^'+19y=0 is y=e^{2t}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))