What is the solution for y^{''}-2y^'+5y=0 ?

The solution for y^{''}-2y^'+5y=0 is y=e^t(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}-2y^'+5y=0

y^{^{''}} - 2 y^{^{'}} + 5 y = 0

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

x \to 0 + lim (cos (x) ln (x))

x \to 0 lim (\frac{sin ( x ) - x}{x sin ( x )})

\frac{\partial}{\partial y} (e^{x^{2} + y^{2}} \cdot 2 x)

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

\int (4 x^{3} - 9 x^{2} + 7 x + 3) e^{- x} d x

What is the solution for y^{''}-2y^'+5y=0 ?
The solution for y^{''}-2y^'+5y=0 is y=e^t(c_{1}cos(2t)+c_{2}sin(2t))