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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{''}} - y^{^{'}} - 3 y = 0

(D^{2} - 3 D) y = 0

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

y^{^{''}} + 2 y^{^{'}} + 2 y = 0, y^{0} = 0, y^{^{'}} 0 = 1

4 \frac{d ^{2} y}{d x ^{2}} + 32 \frac{d y}{d x} + 64 y = 0

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