What is the solution for 1/8 (d^2x)/(dt^2)=-2x-1(dx)/(dt) ?

The solution for 1/8 (d^2x)/(dt^2)=-2x-1(dx)/(dt) is x=c_{1}e^{-4t}+c_{2}te^{-4t}

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1/8 (d^2x)/(dt^2)=-2x-1(dx)/(dt)

\frac{1}{8} \frac{d ^{2} x}{d t ^{2}} = - 2 x - 1 \frac{d x}{d t}

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

Solution steps

\frac{1}{8} \frac{d ^{2} x}{d t ^{2}} = - 2 x - 1 \frac{d x}{d t}

Solve linear ODE:

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

x = c_{1} e^{- 4 t} + c_{2} t e^{- 4 t}

\int (2 x - 4) (x + 2) d x

x - 4, x = 8

\frac{\partial}{\partial x} (e^{x} + z^{2})

2 (y - 4 x^{2}) d x + x d y = 0

\int \frac{1}{cos ^{2} ( x ) + 3 sin ^{2} ( x )} d x

What is the solution for 1/8 (d^2x)/(dt^2)=-2x-1(dx)/(dt) ?
The solution for 1/8 (d^2x)/(dt^2)=-2x-1(dx)/(dt) is x=c_{1}e^{-4t}+c_{2}te^{-4t}