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

The solution for y^{''}+3y=0,y(0)=4,y^'(0)=-3 is y=4cos(sqrt(3)t)-sqrt(3)sin(sqrt(3)t)

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

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

y = 4 cos (3 t) - 3 sin (3 t)

Solution steps

y^{''} (t) + 3 y = 0

Solve linear ODE:

y = 4 cos (3 t) - 3 sin (3 t)

y = 4 cos (3 t) - 3 sin (3 t)

y^{^{''}} - 11 y^{^{'}} + 24 y = 0, y^{^{'}} (0) = 12

\frac{d ^{8} y}{d x ^{8}} - 8 \frac{d ^{4} y}{d x ^{4}} + 16 y = 0

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

y^{^{''}} + 4 y - 3 y^{^{'}} = 0

- 100 y + 65 y^{^{'}} - 14 y^{^{''}} + y^{^{'''}} = 0

What is the solution for y^{''}+3y=0,y(0)=4,y^'(0)=-3 ?
The solution for y^{''}+3y=0,y(0)=4,y^'(0)=-3 is y=4cos(sqrt(3)t)-sqrt(3)sin(sqrt(3)t)