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

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

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

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

y = e^{2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

Solution steps

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

Solve linear ODE:

y = e^{2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y = e^{2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

\frac{\partial}{\partial x} (\frac{1}{x + 2 c t})

x \to 0 lim (\frac{x + 2}{x ^{3} + 8})

lo g_{3} (3 x^{3} + 4 x + 4)

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\int \frac{1}{e ^{2 y} - 2 e ^{y}} d y

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