What is the solution for y^{'''}+10y^{''}+25y^'=0 ?

The solution for y^{'''}+10y^{''}+25y^'=0 is y=c_{1}+c_{2}e^{-5t}+c_{3}te^{-5t}

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y^{'''}+10y^{''}+25y^'=0

y^{^{'''}} + 10 y^{^{''}} + 25 y^{^{'}} = 0

y = c_{1} + c_{2} e^{- 5 t} + c_{3} t e^{- 5 t}

Solution steps

y^{'''} (t) + 10 y^{''} (t) + 25 y^{'} (t) = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 5 t} + c_{3} t e^{- 5 t}

y = c_{1} + c_{2} e^{- 5 t} + c_{3} t e^{- 5 t}

\frac{\partial}{\partial x} (- 2 x y + x^{3})

f (x) = 2 - 3 x^{2}, at x = - 2

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

\frac{d}{d x} (e^{2 x} \cdot sin (x^{2} + 5))

2 x^{3} - x^{2} + 3

What is the solution for y^{'''}+10y^{''}+25y^'=0 ?
The solution for y^{'''}+10y^{''}+25y^'=0 is y=c_{1}+c_{2}e^{-5t}+c_{3}te^{-5t}