What is the solution for (y^{''})/2+(y^')/2+6=0 ?

The solution for (y^{''})/2+(y^')/2+6=0 is y=c_{1}+c_{2}e^{-t}-12t

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(y^{''})/2+(y^')/2+6=0

\frac{y ^{^{''}}}{2} + \frac{y ^{^{'}}}{2} + 6 = 0

y = c_{1} + c_{2} e^{- t} - 12 t

Solution steps

\frac{y ^{''} ( t )}{2} + \frac{y ^{'} ( t )}{2} + 6 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} - 12 t

y = c_{1} + c_{2} e^{- t} - 12 t

y^{^{''}} + 4 y^{^{'}} + 3 y = 5 e^{2} x

y^{^{''}} + 13 y = 7 e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = - 19 t e^{4 t}

2 y^{^{''}} - y^{^{'}} = 7

y^{^{''}} + y^{^{'}} - 6 y = x^{2} - x + e^{2 x}

What is the solution for (y^{''})/2+(y^')/2+6=0 ?
The solution for (y^{''})/2+(y^')/2+6=0 is y=c_{1}+c_{2}e^{-t}-12t