What is the solution for y^{''}+9y^'=40e^{1t} ?

The solution for y^{''}+9y^'=40e^{1t} is y=c_{1}+c_{2}e^{-9t}+4e^t

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y^{''}+9y^'=40e^{1t}

y^{^{''}} + 9 y^{^{'}} = 40 e^{1 t}

y = c_{1} + c_{2} e^{- 9 t} + 4 e^{t}

Solution steps

y^{''} (t) + 9 y^{'} (t) = 40 e^{1 t}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 9 t} + 4 e^{t}

y = c_{1} + c_{2} e^{- 9 t} + 4 e^{t}

\int 5 e^{t} d t

x \to 0 lim (co tan (x))

x \to - 1 lim (x^{3} - 2 x^{2} + 3 x - 4)

f (x) = \frac{3}{x + 1}

x \to 0 lim (2 x + e^{x})

What is the solution for y^{''}+9y^'=40e^{1t} ?
The solution for y^{''}+9y^'=40e^{1t} is y=c_{1}+c_{2}e^{-9t}+4e^t