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

The solution for 2y^{''}+y^'+y=0 is y=e^{-t/4}(c_{1}cos((sqrt(7)t)/4)+c_{2}sin((sqrt(7)t)/4))

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

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

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

Solution steps

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

Solve linear ODE:

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

y = e^{- \frac{t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

t^{2} (5 t + 9)^{3}

2 x^{3} - 3 x^{2}

\frac{1}{x - 1}, 0

x \to 1 lim (\frac{sin ( x )}{x})

\frac{3}{( 1 - x ^{2} ) ^{4}}

What is the solution for 2y^{''}+y^'+y=0 ?
The solution for 2y^{''}+y^'+y=0 is y=e^{-t/4}(c_{1}cos((sqrt(7)t)/4)+c_{2}sin((sqrt(7)t)/4))