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

The solution for y^{''}-y^'+5y=0 is y=e^{t/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))

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

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

y = e^{\frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{19 t}{2}) + c_{2} sin (\frac{19 t}{2}))

t e^{- 2 t}

\frac{\partial}{\partial x} (\frac{y}{x + 1})

\frac{s}{1}

\frac{\partial}{\partial x} (- x^{2} y + \frac{1}{y})

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

What is the solution for y^{''}-y^'+5y=0 ?
The solution for y^{''}-y^'+5y=0 is y=e^{t/2}(c_{1}cos((sqrt(19)t)/2)+c_{2}sin((sqrt(19)t)/2))