What is the solution for 5x^{''}+20x^'+20x=28 ?

The solution for 5x^{''}+20x^'+20x=28 is x=c_{1}e^{-2t}+c_{2}te^{-2t}+7/5

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5x^{''}+20x^'+20x=28

5 x^{^{''}} + 20 x^{^{'}} + 20 x = 28

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7}{5}

Solution steps

5 x^{''} (t) + 20 x^{'} (t) + 20 x = 28

Solve linear ODE:

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7}{5}

x = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{7}{5}

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

\frac{\partial}{\partial x} (5 x^{4} y^{- 4})

\int a^{2 x} d x

\int arcsin (b x) d x

y \frac{d y}{d x} = sin (x)

What is the solution for 5x^{''}+20x^'+20x=28 ?
The solution for 5x^{''}+20x^'+20x=28 is x=c_{1}e^{-2t}+c_{2}te^{-2t}+7/5