What is the solution for 3y^{''}+2y^'+1=0 ?

The solution for 3y^{''}+2y^'+1=0 is y=c_{1}+c_{2}e^{-(2t)/3}-t/2

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3y^{''}+2y^'+1=0

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

y = c_{1} + c_{2} e^{- \frac{2 t}{3}} - \frac{t}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{2 t}{3}} - \frac{t}{2}

y = c_{1} + c_{2} e^{- \frac{2 t}{3}} - \frac{t}{2}

y^{^{''}} - y^{^{'}} - 6 y = e^{- 3 x} (7 - 8 x + 6 x^{2})

y^{^{''}} - 2 y^{^{'}} + y = e^{t}, y (0) = 0, y^{^{'}} (0) = 9

y^{^{''}} - y = x^{2} e x + 3

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

y^{^{''}} - y = x^{2} e x + 5

What is the solution for 3y^{''}+2y^'+1=0 ?
The solution for 3y^{''}+2y^'+1=0 is y=c_{1}+c_{2}e^{-(2t)/3}-t/2