What is the solution for x^{''}+9x=0,x(0)=5,x^'(0)=-11 ?

The solution for x^{''}+9x=0,x(0)=5,x^'(0)=-11 is x=5cos(3t)-11/3 sin(3t)

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x^{''}+9x=0,x(0)=5,x^'(0)=-11

x^{^{''}} + 9 x = 0, x (0) = 5, x^{^{'}} (0) = - 11

x = 5 cos (3 t) - \frac{11}{3} sin (3 t)

Solution steps

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

Solve linear ODE:

x = 5 cos (3 t) - \frac{11}{3} sin (3 t)

x = 5 cos (3 t) - \frac{11}{3} sin (3 t)

y^{^{''''}} + 5 y^{^{'''}} - 2 y^{^{''}} - 3 y^{^{'}} + \frac{1}{2} y = 0

(2 d^{5} - 7 d^{4} + 12 d^{3} + 8 d^{2}) y = 0

y^{^{''}} + y^{^{'}} - 2 y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = - 1, y^{^{'}} (0) = - 3

y^{^{''}} - 5 y^{^{'}} + 4 y = 0, y (0) = \frac{4}{3}, y^{^{'}} (0) = \frac{1}{3}

What is the solution for x^{''}+9x=0,x(0)=5,x^'(0)=-11 ?
The solution for x^{''}+9x=0,x(0)=5,x^'(0)=-11 is x=5cos(3t)-11/3 sin(3t)