What is the solution for 2y^{''}+20y^'+50y=0 ?

The solution for 2y^{''}+20y^'+50y=0 is y=c_{1}e^{-5t}+c_{2}te^{-5t}

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2y^{''}+20y^'+50y=0

2 y^{^{''}} + 20 y^{^{'}} + 50 y = 0

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

Solution steps

2 y^{''} (t) + 20 y^{'} (t) + 50 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (2 x^{2} + y^{2})

\frac{d}{d x} (x^{25})

∣ x ∣ + 1

x \to 0 lim (4 \cdot x^{2})

\int 1 + cos (2 x) d x

What is the solution for 2y^{''}+20y^'+50y=0 ?
The solution for 2y^{''}+20y^'+50y=0 is y=c_{1}e^{-5t}+c_{2}te^{-5t}