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

The solution for y^{''}+4y=0,y(0)=3,y^'(0)=-3 is y=3cos(2t)-3/2 sin(2t)

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

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

y = 3 cos (2 t) - \frac{3}{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = 3 cos (2 t) - \frac{3}{2} sin (2 t)

y = 3 cos (2 t) - \frac{3}{2} sin (2 t)

5 y^{^{''}} + 5 y^{^{'}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

y^{^{''}} - \frac{10}{3} y^{^{'}} + α y = 0, y^{^{'}} (0) = - \frac{2}{3}

x^{^{''}} - x^{^{'}} - \frac{3}{4} x = 0

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

y^{^{''}} + 8 y^{^{'}} + 17 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

What is the solution for y^{''}+4y=0,y(0)=3,y^'(0)=-3 ?
The solution for y^{''}+4y=0,y(0)=3,y^'(0)=-3 is y=3cos(2t)-3/2 sin(2t)