What is the solution for y^{''}+1/2 y^'+y=0 ?

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

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y^{''}+1/2 y^'+y=0

y^{^{''}} + \frac{1}{2} y^{^{'}} + y = 0

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

Solution steps

y^{''} (t) + \frac{1}{2} y^{'} (t) + y = 0

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{15 t}{4}) + c_{2} sin (\frac{15 t}{4}))

\int \frac{ln ( 3 x )}{x} d x

x^{2} + 3 x - 4

x^{2} - 3 x

x \to - \infty lim (2 x^{3})

x \to 0 lim ((1 + 8 x)^{\frac{7}{x}})

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