What is the solution for y^{''}+6y^'+8=0 ?

The solution for y^{''}+6y^'+8=0 is y=c_{1}+c_{2}e^{-6t}-(4t)/3

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y^{''}+6y^'+8=0

y^{^{''}} + 6 y^{^{'}} + 8 = 0

y = c_{1} + c_{2} e^{- 6 t} - \frac{4 t}{3}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 8 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 6 t} - \frac{4 t}{3}

y = c_{1} + c_{2} e^{- 6 t} - \frac{4 t}{3}

\frac{d y}{d x} = 2 x (x - 4)

\frac{d y}{d x} - \frac{y}{x} = 2 x e^{x}

n = 1 \sum \infty (\frac{2 x + 1}{x})^{n}

\int_{- 48}^{0} \frac{1}{1 - x} d x

\int \frac{cos ( 2 x )}{sin ^{3} ( 2 x )} d x

What is the solution for y^{''}+6y^'+8=0 ?
The solution for y^{''}+6y^'+8=0 is y=c_{1}+c_{2}e^{-6t}-(4t)/3