What is the solution for x^{''}+4x=0,x(0)=1,x^'(0)=-2 ?

The solution for x^{''}+4x=0,x(0)=1,x^'(0)=-2 is x=cos(2t)-sin(2t)

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x^{''}+4x=0,x(0)=1,x^'(0)=-2

x^{^{''}} + 4 x = 0, x (0) = 1, x^{^{'}} (0) = - 2

x = cos (2 t) - sin (2 t)

Solution steps

x^{''} (t) + 4 x = 0

Solve linear ODE:

x = cos (2 t) - sin (2 t)

x = cos (2 t) - sin (2 t)

y^{^{''}} + 12 y^{^{'}} + 37 y = 4 y

y^{^{''}} - y^{^{'}} - 12 y = 0, y (1) = 0 y^{^{'}} (1)

y^{^{'''}} - 5 y^{^{''}} - 4 y^{^{'}} + 20 y = 0

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = - 3

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0, y = - cos (2 x)

What is the solution for x^{''}+4x=0,x(0)=1,x^'(0)=-2 ?
The solution for x^{''}+4x=0,x(0)=1,x^'(0)=-2 is x=cos(2t)-sin(2t)