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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1 + x}{x} d x

n = 1 \sum \infty \frac{2 n}{5 n - 4}

\int \frac{x ^{2} - 4}{x ^{6}} d x

y = x^{3} - 3 x^{2} + 25

\int_{0}^{π} 9 e^{x} sin (x) 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=cos(3t)+2sin(3t)+3