What is the solution for y^{''}-y^'+12y=0 ?

The solution for y^{''}-y^'+12y=0 is y=e^{t/2}(c_{1}cos((sqrt(47)t)/2)+c_{2}sin((sqrt(47)t)/2))

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y^{''}-y^'+12y=0

y^{^{''}} - y^{^{'}} + 12 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2}))

Solution steps

y^{''} (t) - y^{'} (t) + 12 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{47 t}{2}) + c_{2} sin (\frac{47 t}{2}))

\frac{d}{d x} (\frac{( e ^{x} )}{1 - e ^{- x}})

\int - \frac{x}{x ^{2} + y ^{2}} d x

y^{^{'}} + \frac{y}{t} = 2 e^{t}

x^{2} + 6 x

y^{^{'}} = \frac{2}{3 x} y + \frac{2 x}{y}

What is the solution for y^{''}-y^'+12y=0 ?
The solution for y^{''}-y^'+12y=0 is y=e^{t/2}(c_{1}cos((sqrt(47)t)/2)+c_{2}sin((sqrt(47)t)/2))