What is the solution for y^{''}+10y^'+25y=4e^{-5t} ?

The solution for y^{''}+10y^'+25y=4e^{-5t} is y=c_{1}e^{-5t}+c_{2}te^{-5t}+2t^2e^{-5t}

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y^{''}+10y^'+25y=4e^{-5t}

y^{^{''}} + 10 y^{^{'}} + 25 y = 4 e^{- 5 t}

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

Solution steps

y^{''} (t) + 10 y^{'} (t) + 25 y = 4 e^{- 5 t}

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (13.12 + 0.6215 x - 11.37 y^{0.16} + 0.3965 x y^{0.16})

f (x) = sin (x) cos (x)

\int \frac{( a + b x ^{9} )}{10 a x + b x ^{10}} d x

\int \frac{x}{( x ^{2} + 1 ) ( x ^{2} + 4 x + 4 )} d x

x \to 0 lim (8 x csc (5 x))

What is the solution for y^{''}+10y^'+25y=4e^{-5t} ?
The solution for y^{''}+10y^'+25y=4e^{-5t} is y=c_{1}e^{-5t}+c_{2}te^{-5t}+2t^2e^{-5t}