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

The solution for y^{''}+y^'+4y=0 is y=e^{-t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))

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

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

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{15 t}{2}) + c_{2} sin (\frac{15 t}{2}))

6 x sin (x)

\frac{\partial}{\partial y} (x^{\frac{x}{y}} cos (\frac{x}{y}))

((4 - x^{\frac{2}{3}})^{\frac{3}{2}})^{^{'}}

f (x) = \frac{1 + 2 x}{3 + x}

f (x) = (x - 1) e^{x^{2} - 4}

What is the solution for y^{''}+y^'+4y=0 ?
The solution for y^{''}+y^'+4y=0 is y=e^{-t/2}(c_{1}cos((sqrt(15)t)/2)+c_{2}sin((sqrt(15)t)/2))