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

The solution for x^{''}+9x-27=0,x(0)=4,x^'(0)=6 is x=cos(3t)+2sin(3t)+3

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

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

x = cos (3 t) + 2 sin (3 t) + 3

Solution steps

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

Solve linear ODE:

x = cos (3 t) + 2 sin (3 t) + 3

x = cos (3 t) + 2 sin (3 t) + 3

\int \frac{1}{12 y} d y

\frac{\partial}{\partial y} (\frac{y - t}{5 y + 2 t})

t - π

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

\frac{d}{d x} (ln (x^{2} - 4))

What is the solution for x^{''}+9x-27=0,x(0)=4,x^'(0)=6 ?
The solution for x^{''}+9x-27=0,x(0)=4,x^'(0)=6 is x=cos(3t)+2sin(3t)+3