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

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

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

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

y = - cos (2 t)

Solution steps

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

Solve linear ODE:

y = - cos (2 t)

y = - cos (2 t)

y^{^{'''}} + 6 y^{^{''}} + 11 y^{^{'}} + 6 y = 0

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + y = 0

y^{^{''}} - 6 y^{^{'}} + 34 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

y^{^{''}} - 10 y^{^{'}} + 50 y = 0, y (0) = - 4, y^{^{'}} (0) = - 5

9 y^{^{''}} - 12 y^{^{'}} + 4 y = 0, y (0) = e^{(\frac{2 x}{3})}

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