What is the solution for x^{''}+9x=0,x^'(0)=4,x(0)=3 ?

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

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x^{''}+9x=0,x^'(0)=4,x(0)=3

x^{^{''}} + 9 x = 0, x^{^{'}} (0) = 4, x (0) = 3

x = 3 cos (3 t) + \frac{4}{3} sin (3 t)

Solution steps

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

Solve linear ODE:

x = 3 cos (3 t) + \frac{4}{3} sin (3 t)

x = 3 cos (3 t) + \frac{4}{3} sin (3 t)

y^{^{''}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 22

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 15 y = 0

y^{^{''}} - 19 y^{^{'}} - 10 y = 0, y (0) = 1, y^{^{'}} (0) = 1

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

y^{^{''}} + 40 y^{^{'}} + 4000 y = 0

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