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

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

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

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

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^{^{''}} - 7 y^{^{'}} + 2 y = 0

\frac{d ^{2} y}{d t ^{2}} + 7 \frac{d y}{d t} + k y = 0

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

x, \frac{d ^{2} x}{d y ^{2}} + 0.1 x = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

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