What is the solution for x^{''}+144x=0,x(0)=3,x^'(0)=-48 ?

The solution for x^{''}+144x=0,x(0)=3,x^'(0)=-48 is x=3cos(12t)-4sin(12t)

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x^{''}+144x=0,x(0)=3,x^'(0)=-48

x^{^{''}} + 144 x = 0, x (0) = 3, x^{^{'}} (0) = - 48

x = 3 cos (12 t) - 4 sin (12 t)

Solution steps

x^{''} (t) + 144 x = 0

Solve linear ODE:

x = 3 cos (12 t) - 4 sin (12 t)

x = 3 cos (12 t) - 4 sin (12 t)

What is the solution for x^{''}+144x=0,x(0)=3,x^'(0)=-48 ?
The solution for x^{''}+144x=0,x(0)=3,x^'(0)=-48 is x=3cos(12t)-4sin(12t)