What is the solution for x^{''}+9x-27=0 ?

The solution for x^{''}+9x-27=0 is x=c_{1}cos(3t)+c_{2}sin(3t)+3

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x^{''}+9x-27=0

x^{^{''}} + 9 x - 27 = 0

x = c_{1} cos (3 t) + c_{2} sin (3 t) + 3

Solution steps

x^{''} (t) + 9 x - 27 = 0

Solve linear ODE:

x = c_{1} cos (3 t) + c_{2} sin (3 t) + 3

x = c_{1} cos (3 t) + c_{2} sin (3 t) + 3

\int_{3}^{5} \frac{2}{4 + ( x - 3 ) ^{2}} d x

\frac{d}{d x} ((2 x - 1) (3 - 4 x))

\frac{d}{d s} (\frac{1}{s + 2})

\int \frac{2}{1 - sin ( x )} d x

\int - 0.01 x d x

What is the solution for x^{''}+9x-27=0 ?
The solution for x^{''}+9x-27=0 is x=c_{1}cos(3t)+c_{2}sin(3t)+3