What is the solution for x^{''}+4x^'=-2x^'-8x+3x ?

The solution for x^{''}+4x^'=-2x^'-8x+3x is x=c_{1}e^{-t}+c_{2}e^{-5t}

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x^{''}+4x^'=-2x^'-8x+3x

x^{^{''}} + 4 x^{^{'}} = - 2 x^{^{'}} - 8 x + 3 x

x = c_{1} e^{- t} + c_{2} e^{- 5 t}

Solution steps

x^{''} (t) + 4 x^{'} (t) = - 2 x^{'} (t) - 8 x + 3 x

Solve linear ODE:

x = c_{1} e^{- t} + c_{2} e^{- 5 t}

x = c_{1} e^{- t} + c_{2} e^{- 5 t}

12 y^{^{''}} - 5 y - 2 y = 0

y^{^{''}} + 2 y = 0, y (0) = sin (k t)

\frac{d ^{2} y}{d t ^{2}} = 5 \frac{d y}{d t} - 4 y

f^{^{''}} (x) - f^{^{'}} (x) - 2 f (x) = 0, f (0) = - 2, f^{^{'}} (0) = 2

9 y^{^{''}} - 12 y^{^{'}} + 11 y = 0

What is the solution for x^{''}+4x^'=-2x^'-8x+3x ?
The solution for x^{''}+4x^'=-2x^'-8x+3x is x=c_{1}e^{-t}+c_{2}e^{-5t}