What is the solution for y^{''}+3y^'+4=0 ?

The solution for y^{''}+3y^'+4=0 is y=c_{1}+c_{2}e^{-3t}-(4t)/3

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y^{''}+3y^'+4=0

y^{^{''}} + 3 y^{^{'}} + 4 = 0

y = c_{1} + c_{2} e^{- 3 t} - \frac{4 t}{3}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 4 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} - \frac{4 t}{3}

y = c_{1} + c_{2} e^{- 3 t} - \frac{4 t}{3}

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{x} )}{x}

y^{^{''}} + 9 y = 3 sec^{3} (3 x)

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

y^{^{''}} + (- 3) y^{^{'}} + 4 y = 16 x - 12 e^{2 x}

y^{^{''''}} - 3 y^{^{'''}} + 2 y^{^{''}} = 4 x

What is the solution for y^{''}+3y^'+4=0 ?
The solution for y^{''}+3y^'+4=0 is y=c_{1}+c_{2}e^{-3t}-(4t)/3