What is the solution for x^{''}+9x^'=27,x(0)=4,x^'(0)=6 ?

The solution for x^{''}+9x^'=27,x(0)=4,x^'(0)=6 is x=4+1/3-1/3 e^{-9t}+3t

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x^{''}+9x^'=27,x(0)=4,x^'(0)=6

x^{^{''}} + 9 x^{^{'}} = 27, x (0) = 4, x^{^{'}} (0) = 6

x = 4 + \frac{1}{3} - \frac{1}{3} e^{- 9 t} + 3 t

Solution steps

x^{''} (t) + 9 x^{'} (t) = 27

Solve linear ODE:

x = 4 + \frac{1}{3} - \frac{1}{3} e^{- 9 t} + 3 t

x = 4 + \frac{1}{3} - \frac{1}{3} e^{- 9 t} + 3 t

x \to 0 lim (\frac{cos ( 1 )}{x})

\frac{d}{d x} (a ln (x^{2}))

\int 7 x^{3} d x

f (x) = - 5 x^{2} + x, at x = 3

\int_{0}^{1} (x (x + 1)) d x

What is the solution for x^{''}+9x^'=27,x(0)=4,x^'(0)=6 ?
The solution for x^{''}+9x^'=27,x(0)=4,x^'(0)=6 is x=4+1/3-1/3 e^{-9t}+3t