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

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

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

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

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

Solution steps

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

e^{x^{2} - 4 x}

8 e^{4 x}

\frac{d}{d x} (tan (x) - sec (x))

\frac{\partial}{\partial x} (x^{3} + y^{3} - 9 x y)

x \to \frac{π}{4} lim (tan (x))

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